Understanding Motion in Two Dimensions - High School Physics

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Question

Peter walks $30.7m \text{ at }38.3^{\circ}$ north of east. What is his total displacement along the y-axis?

Answer

Start by drawing a picture.

By traveling both north and east, Peter has displacement along both the x-axis and the y-axis. Recognize that this makes a right triangle.

We can solve for the y-axis displacement by using trigonometry.

$\sin(\Theta)=\frac{\text{opposite}}{\text{hypotenuse}}$

In this case, we know the angle and the hypotenuse. The y-displacement is opposite the angle.

$\sin(38.3^o)=\frac{Y}{30.7m}$

Multiply both sides by and solve.

$(30.7m)\sin(38.3^o)=Y$

Since we are solving for the displacement, we need to include the direction. The displacement is $\small 19.03m\ \text{north}$ .

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Question

Peter walks $30.7m \text{ at }38.3^{\circ}$ north of east. What is his total displacement along the x-axis?

Answer

Start by drawing a picture.

By traveling both north and east, Peter has displacement along both the x-axis and the y-axis. Recognize that this makes a right triangle.

We can solve for the x-axis displacement by using trigonometry.

$\cos(\Theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$

In this case, we know the angle and the hypotenuse. The x-displacement is adjacent to the angle.

$\cos(38.3^o)=\frac{X}{30.7m}$

Multiply both sides by and solve.

$(30.7m)\cos(38.3^o)=X$

Since we are solving for the displacement, we need to include the direction. The displacement is $\small 24.09m\ \text{east}$ .

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

At what height above the ground will the rock change direction?

Answer

Even though the problem gives us an initial velocity, we need to break it down into horizontal and vertical components.

$v_{i} \sin(\Theta)=v_{iy}$

We can plug in the given values and find the vertical velocity.

$30\frac{m}{s} \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

Remember that the vertical velocity at the highest point of a parabola is zero. Now that we know the initial and final vertical velocities, we can plug our values into an equation to solve for the maximum height.

$v_{f}^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(296.09\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(296.09\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-296.09\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$15.11m=\Delta y$

Remember, $\Delta y$ only tells us the CHANGE in the vertical direction. The rock started at the top of a tall building, then rose an extra .

Its highest point is above the ground.

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

How long is the rock in the air?

Answer

We first need to find the vertical component of the velocity.

$v_{i}\sin(\Theta)=v_{iy}$

We can plug in the given values for the angle and the initial velocity to find the vertical component.

$30\frac{m}{s} \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

Now we need to solve for the time that the rock travels upward. We can then add the upward travel time to the downward travel time to find the total time in the air.

Remember that the vertical velocity at the highest point of a parabola is zero. We can use that to find the time for the rock to travel upward.

$0\frac{m}{s}=17.21\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-17.21\frac{m}{s}=(-9.8\frac{m}{s^2}) t_1$

$\frac{-17.21\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

Now let's find the time for the downward travel. We don't know the final velocity for the rock, but we CAN use the information we have been given to find the height it travels upward.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(296.18\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(296.18\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-296.18\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$15.11m=\Delta y$

Remember, $\Delta y$ only tells us the vertical CHANGE. Since the rock started at the top of a building, if it rose an extra , then at its highest point it is above the ground.

This means that our $\Delta y$ will be as it will be traveling down from the highest point. Using this distance, we can find the downward travel time.

$\Delta y =v_it+\frac{1}{2}at^2$

$-25.11m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=(-4.9\frac{m}{s^2})t^2$

$\frac{-25.11m}{-4.9\frac{m}{s^2}}=t^2$

$\sqrt{5.12s^2}=\sqrt{t^2}$

Add together the time for upward travel and downward travel to find the total flight time.

$t_{total}=t_1+t_2$

$t_{total}=1.76s+2.26s$

$t_{total}=4.02s$

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

What is the horizontal distance that the rock travels?

Answer

We first need to find the horizontal component of the initial velocity.

$v_{i}* \cos(\Theta)=v_{x}$

We can plug in the given values for the angle and initial velocity and solve.

$30\frac{m}{s}* \cos(35^{\circ})=v_{x}$

$30\frac{m}{s}* (0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

The only force acting on the rock during flight is gravity; there are no forces in the horizontal direction, meaning that the horizontal velocity will remain constant. We can set up a simple equation to find the relationship between distance traveled and the velocity.

$\Delta x =v_xt$

We know , but now we need to find the time the rock is in the air.

We need to solve for the time that the rock travels upward. We can then add the upward travel time to the downward travel time to find the total time in the air.

Remember that the vertical velocity at the highest point of a parabola is zero. We can use that to find the time for the rock to travel upward.

$0\frac{m}{s}=17.21\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-17.21\frac{m}{s}=(-9.8\frac{m}{s^2}) t_1$

$\frac{-17.21\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

Now let's find the time for the downward travel. We don't know the final velocity for the rock, but we CAN use the information we have been given to find the height it travels upward.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(296.18\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(296.18\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-296.18\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$15.11m=\Delta y$

Remember, $\Delta y$ only tells us the vertical CHANGE. Since the rock started at the top of a building, if it rose an extra , then at its highest point it is above the ground.

This means that our $\Delta y$ will be as it will be traveling down from the highest point. Using this distance, we can find the downward travel time.

$\Delta y =v_it+\frac{1}{2}at^2$

$-25.11m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=(-4.9\frac{m}{s^2})t^2$

$\frac{-25.11m}{-4.9\frac{m}{s^2}}=t^2$

$\sqrt{5.12s^2}=\sqrt{t^2}$

Add together the time for upward travel and downward travel to find the total flight time.

$t_{total}=t_1+t_2$

$t_{total}=1.76s+2.26s$

$t_{total}=4.02s$

Now that we've finally found our time, we can plug that back into the equation from the beginning of the problem, along with our horizontal velocity, to solve for the final distance.

$\Delta x =v_xt$

$\Delta x =24.57\frac{m}{s}* 4.02s$

$\Delta x=98.77m$

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

What is the net final velocity of the rock right before it hits the ground?

Answer

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation $v_{i}\cos(\Theta)=v_{x}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}\cos(35^{\circ})=v_{x}$

$30\frac{m}{s}(0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

We can find the vertical velocity using the equation $v_{i} \sin(\Theta)=v_{iy}$ .

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

The horizontal velocity will not change during flight because there are no forces in the horizontal direction. The vertical velocity, however, will be affected. We need to solve for the final vertical velocity, then combine the vertical and horizontal vectors to find the total final velocity.

We know that the rock is going to travel a net distance of , as that is the distance between where the rock's initial and final positions. We now know the displacement, initial velocity, and acceleration, which will allow us to solve for the final velocity.

$v_f^2=v_1^2+2a\Delta y$

$v_f^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)$

$v_f^2=(296.18\frac{m^2}{s^2})+196\frac{m^2}{s^2}$

$v_f^2=492.18\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{492.18\frac{m^2}{s^2}}$

$v_f=22.19\frac{m}{s}$

Because the rock is traveling downward, our velocity will be negative: $v_f=-22.19\frac{m}{s}$ .

Now that we know our final velocities in both the horizontal and vertical directions, we can use the Pythagorean theorem to solve for the net velocity.

$(-22.19\frac{m}{s})^2+(24.57\frac{m}{s})^2=(v_{resultant})^2$

$492.4\frac{m^2}{s^2}+603.68\frac{m^2}{s^2}=(v_{resultant})^2$

$1096.08\frac{m^2}{s^2}=(v_{resultant})^2$

$\sqrt{1096.08\frac{m^2}{s^2}}=\sqrt{(v_{resultant})^2}$

$33.1\frac{m}{s}=v_{resultant}$

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

At what angle to the horizontal will the rock impact the ground?

Answer

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation $v_{i}\cos(\Theta)=v_{x}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}\cos(35^{\circ})=v_{x}$

$30\frac{m}{s}(0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

We can find the vertical velocity using the equation $v_{i} \sin(\Theta)=v_{iy}$ .

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

The horizontal velocity will not change during flight because there are no forces in the horizontal direction. The vertical velocity, however, will be affected. We need to solve for the final vertical velocity, then combine the vertical and horizontal vectors to find the total final velocity.

We know that the rock is going to travel a net distance of , as that is the distance between where the rock's initial and final positions. We now know the displacement, initial velocity, and acceleration, which will allow us to solve for the final velocity.

$v_f^2=v_1^2+2a\Delta y$

$v_f^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)$

$v_f^2=(296.18\frac{m^2}{s^2})+196\frac{m^2}{s^2}$

$v_f^2=492.18\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{492.18\frac{m^2}{s^2}}$

$v_f=22.19\frac{m}{s}$

Because the rock is traveling downward, our velocity will be negative: $v_f=-22.19\frac{m}{s}$ .

Now that we know our final velocities in both the horizontal and vertical directions, we can find the angle created between the two trajectories. The horizontal and vertical velocities can be compared using trigonometry.

$\tan(\Theta)=\frac{v_y}{v_x}$ ,

$\tan^{-1}(\frac{v_y}{v_x})=\Theta$

Plug in our values and solve for the angle.

$\tan^{-1}(\frac{22.19\frac{m}{s}}{24.57\frac{m}{s}})=\Theta$

$\tan^{-1}(0.90)=\Theta$

$41.99^{\circ}=\Theta$

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

What is the total velocity of the rock when it reaches maximum height?

Answer

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation $v_{i}\cos(\Theta)=v_{x}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}\cos(35^{\circ})=v_{x}$

$30\frac{m}{s}(0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

We can find the vertical velocity using the equation $v_{i} \sin(\Theta)=v_{iy}$ .

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

The question asks for the total velocity at the maximum height. At the top of the parabola, $v_y=0\frac{m}{s}$ as the direction of the motion changes from upward to downward. Yet, even though it has no vertical velocity, the horizontal velocity remains constant during flight. The only force on the rock is that of gravity, and gravity will only affect the vertical velocity. There are no horizontal forces present to alter the horizontal velocity.

Our final answer will be equal to the horizontal velocity: $24.57\frac{m}{s}$ .

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Question

Laurence throws a rock off the edge of a tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the initial vertical velocity?

Answer

The given initial velocity is at an angle, so we have to use some trigonometric functions to break it into horizontal and vertical components.

Effectively, the initial velocity becomes the hypotenuse of a right triangle with the horizontal velocity becoming the base and the vertical velocity becoming the height. To find the vertical velocity, we use the relationship between the hypotenuse and the opposite side.

$\sin{\theta}=\frac{\text{opposite}}{\text{hypotenuse}}$

$\sin{\theta}=\frac{v_y}{v_{i}}$

$v_{i}\sin(\theta)=v_{y}$

Use the given initial velocity and angle to solve for the vertical velocity.

$31\frac{m}{s}\sin(20^{\circ})=v_{y}$

$31\frac{m}{s}*(0.34)=v_{y}$

$10.6 \frac{m}{s}=v_{y}$

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Question

Laurence throws a rock off the edge of a tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the initial horizontal velocity?

Answer

The given initial velocity is at an angle, so we have to use some trigonometric functions to break it into horizontal and vertical components.

Effectively, the initial velocity becomes the hypotenuse of a right triangle with the horizontal velocity becoming the base and the vertical velocity becoming the height. To find the horizontal velocity, we use the relationship between the hypotenuse and the adjacent side.

$\cos{\theta}=\frac{\text{adjacent}}{\text{hypotenuse}}$

$\cos{\theta}=\frac{v_x}{v_{i}}$

$v_{i}\cdot \cos(\theta)=v_{x}$

Use the given initial velocity and angle to find the horizontal velocity.

$31\frac{m}{s}\cos(20^{\circ})=v_{x}$

$31\frac{m}{s}(0.94)=v_{x}$

$29.13\frac{m}{s}=v_{x}$

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Question

Laurence throws a rock off the edge of a tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the total velocity at the maximum height?

Answer

In a problem with parabolic motion, your first step should always be to break the given velocity into its horizontal and vertical components.

Use cosine to find the initial horizontal velocity.

$v_{i}\cdot \cos(\theta)=v_{x}$

We are given the initial velocity and angle.

$31\frac{m}{s}\cos(20^{\circ})=v_{x}$

$31\frac{m}{s}(0.94)=v_{x}$

$29.13\frac{m}{s}=v_{x}$

Use these values with sine to find the initial vertical velocity.

$v_{i}\sin(\theta)=v_{iy}$

$31\frac{m}{s}\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s}*(0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

The question is asking for the velocity at the maximum height or, in our terms, the top of the parabola. At the top of the parabola the vertical velocity will be zero, but the horizontal velocity will remain constant. To find the total velocity, we normally use the Pythagorean Theorem with the horizontal and vertical velocities.

$v_{R}^2=v_x^2+v_y^2$

Since we know that the vertical velocity is zero, however, we can see that the total velocity will simply be equal to the horizontal velocity.

$\sqrt{v_{R}^2}=\sqrt{(29.13\frac{m}{s})^2}$

$v_{R}=29.13\frac{m}{s}$

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Question

Laurence throws a rock off the edge of a tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

How high will the rock go?

Answer

Even though the problem gives us an initial velocity, we need to break it down into horizontal and vertical components.

Use the sine function with the initial velocity and angle to find the vertical velocity.

$v_{i}\sin(\theta)=v_{iy}$

$31\frac{m}{s}\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s}*(0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

Remember that the vertical velocity at the highest point of a parabola is zero. We know the initial and final velocities, and the acceleration of gravity. Using these values in the appropriate motion equation, we can find the height that the rock travels.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(112.36\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(112.36\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-112.36\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$5.73m=\Delta y$

Remember, $\Delta y$ only tells us the CHANGE in the vertical direction. Since the rock started at the top of a building, it rose an extra after it was thrown. Its highest point it is above the ground.

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Question

Laurence throws a rock off the edge of a tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

How long is the rock in the air?

Answer

The given velocity won't help us much here. We need to break it down into horizontal and vertical components.

Use the sine function with the initial velocity and angle to find the vertical velocity.

$v_{i}\sin(\theta)=v_{iy}$

$31\frac{m}{s}\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s}* (0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

Now we need to work on finding the time in the air. To do this, we need to break the rock's path into two parts. The first part is the time that the rock is rising to its maximum height, and the second part is the time that it is falling from the highest point to the ground.

For the first part, we can assume that the final vertical velocity is zero, since this will be the top of the parabola. Using the initial velocity, final velocity, and gravity, we can solve for the time to travel this portion of the path.

$0\frac{m}{s}=10.6\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-10.6\frac{m}{s}=-9.8\frac{m}{s^2}\cdot t_1$

$\frac{-10.6\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

.

This is the time that the rock is traveling upward. Now we need to focus on the time that the rock travels downward. We don't know the final velocity at the end of the parabola, so we can't use the equation from the first part. If we can find the total height of the parabola, we can use a different equation to solve for time.

$\Delta y =v_it+\frac{1}{2}at^2$

Remember that the vertical velocity at the highest point of a parabola is zero. Now that we know the initial vertical velocity, we can plug this into an equation to solve for the distance that the rock travels from its maximum height to the original height.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(112.36\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(112.36\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-112.36\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$5.73m=\Delta y$

Remember, $\Delta y$ only tells us the CHANGE in height. Since the rock started at the top of a tall building, if it rose an extra , then at its highest point it is above the ground.

This means that our $\Delta y$ for the second equation will be (the change in height is negative because it travels downward.) Use this total distance and the velocity at the top of the peak (zero) to solve for the time that the rock travels down.

$\Delta y =v_it+\frac{1}{2}at^2$

$-17.73m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-17.73m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-17.73m=(-4.9\frac{m}{s^2})t^2$

$\frac{-17.73m}{-4.9\frac{m}{s^2}}=t^2$

$\sqrt{3.62s^2}=\sqrt{t^2}$

Finally, add the two times together to find the total time in flight.

$t_{total}=t_1+t_2$

$t_{total}=1.08s+1.90s$

$t_{total}=2.98s$

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Question

Laurence throws a rock off the edge of a tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the total horizontal distance that the rock travels?

Answer

To find the horizontal distance, we will need to use the horizontal velocity and the total time that the rock is in the air. We can easily find the horizontal velocity by using the cosine function and the initial velocity and angle.

$v_{i}* \cos(\theta)=v_{x}$

$31\frac{m}{s}* \cos(20^{\circ})=v_{x}$

$31\frac{m}{s}(0.94)=v_{x}$

$29.13\frac{m}{s}=v_{x}$

There is no horizontal acceleration, so the velocity will be constant. This means our equation for distance will be simply:

$\Delta x =v_xt$ .

We know , but now we need to find the time the rock is in the air.

Use the sine function with the initial velocity and angle to find the vertical velocity.

$v_{i}\sin(\theta)=v_{iy}$

$31\frac{m}{s}\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s} (0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

We need to break the rock's path into two parts. The first part is the time that the rock is rising to its maximum height, and the second part is the time that it is falling from the highest point to the ground.

For the first part, we can assume that the final vertical velocity is zero, since this will be the top of the parabola. Using the initial velocity, final velocity, and gravity, we can solve for the time to travel this portion of the path.

$0\frac{m}{s}=10.6\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-10.6\frac{m}{s}=-9.8\frac{m}{s^2}\cdot t_1$

$\frac{-10.6\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

.

This is the time that the rock is traveling upward. Now we need to focus on the time that the rock travels downward. We don't know the final velocity at the end of the parabola, so we can't use the equation from the first part. If we can find the total height of the parabola, we can use a different equation to solve for time.

$\Delta y =v_it+\frac{1}{2}at^2$

Remember that the vertical velocity at the highest point of a parabola is zero. Now that we know the initial vertical velocity, we can plug this into an equation to solve for the distance that the rock travels from its maximum height to the original height.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(112.36\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(112.36\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-112.36\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$5.73m=\Delta y$

Remember, $\Delta y$ only tells us the CHANGE in height. Since the rock started at the top of a tall building, if it rose an extra , then at its highest point it is above the ground.

This means that our $\Delta y$ for the second equation will be (the change in height is negative because it travels downward.) Use this total distance and the velocity at the top of the peak (zero) to solve for the time that the rock travels down.

$\Delta y =v_it+\frac{1}{2}at^2$

$-17.73m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-17.73m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-17.73m=(-4.9\frac{m}{s^2})t^2$

$\frac{-17.73m}{-4.9\frac{m}{s^2}}=t^2$

$\sqrt{3.62s^2}=\sqrt{t^2}$

Finally, add the two times together to find the total time in flight.

$t_{total}=t_1+t_2$

$t_{total}=1.08s+1.90s$

$t_{total}=2.98s$

Now that we have finally found our time, we can plug that back into the equation from the beginning of the problem:

$\Delta x =v_xt$

Use the time and horizontal velocity to find the total horizontal distance.

$\Delta x =29.13\frac{m}{s}*2.98s$

$\Delta x=86.81m$

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Question

Laurence throws a rock off the edge of a tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the total final velocity of the rock right as it hits the ground?

Answer

First, break our given velocity into its horizontal and vertical components.

We can find the horizontal velocity using the cosine function, initial velocity, and angle.

$v_{i}\cos(\theta)=v_{x}$

$31\frac{m}{s}\cos(20^{\circ})=v_{x}$

$31\frac{m}{s}(0.94)=v_{x}$

$29.13\frac{m}{s}=v_{x}$

We can find the vertical velocity using the sine function, initial velocity, and angle.

$v_{i} \sin(\theta)=v_{iy}$

$31\frac{m}{s}\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s} (0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

The horizontal velocity is not going to change, since there is no horizontal acceleration. The vertical velocity, however, will be affected by gravity.

We know that the rock is going to travel a net displacement of , as that's the distance between the rock's initial position (the top of the building) and the ground. The displacement must be negative because the rock is traveling downward. Using the appropriate motion equation, we can use the rock's initial velocity, displacement, and acceleration to find the final velocity.

$v_f^2=v_i^2+2a\Delta y$

$v_f^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-12m)$

$v_f^2=(112.36\frac{m^2}{s^2})+235.2\frac{m^2}{s^2}$

$v_f^2=347.56\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{347.56\frac{m^2}{s^2}}$

$v_f=18.64\frac{m}{s}$

Keep in mind that square roots can be positive or negative. Since the rock is traveling in the downward direction, we need to take the negative root.

$v_f=-18.64\frac{m}{s}$ .

We now have both the final horizontal and final vertical velocities. Use the Pythagorean Theorem to find the total velocity from the directional velocities.

$(-18.64\frac{m}{s})^2+(29.13\frac{m}{s})^2=(v_{R})^2$

$347.56\frac{m^2}{s^2}+848.56\frac{m^2}{s^2}=(v_{R})^2$

$1196.2\frac{m^2}{s^2}=(v_{R})^2$

$\sqrt{1196.2\frac{m^2}{s^2}}=\sqrt{(v_{R})^2}$

$34.58\frac{m}{s}=v_{R}$

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Question

Laurence throws a rock off the edge of a tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the angle that the rock will make with the ground right before it hits the ground?

Answer

To find the angle, we will need to find the final horizontal and vertical velocities. First, break our given velocity into its horizontal and vertical components.

We can find the horizontal velocity using the cosine function, initial velocity, and angle.

$v_{i}\cos(\theta)=v_{x}$

$31\frac{m}{s}\cos(20^{\circ})=v_{x}$

$31\frac{m}{s}(0.94)=v_{x}$

$29.13\frac{m}{s}=v_{x}$

We can find the vertical velocity using the sine function, initial velocity, and angle.

$v_{i} \sin(\theta)=v_{iy}$

$31\frac{m}{s}\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s} (0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

The horizontal velocity is not going to change, since there is no horizontal acceleration. The vertical velocity, however, will be affected by gravity.

We know that the rock is going to travel a net displacement of , as that's the distance between the rock's initial position (the top of the building) and the ground. The displacement must be negative because the rock is traveling downward. Using the appropriate motion equation, we can use the rock's initial velocity, displacement, and acceleration to find the final velocity.

$v_f^2=v_i^2+2a\Delta y$

$v_f^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-12m)$

$v_f^2=(112.36\frac{m^2}{s^2})+235.2\frac{m^2}{s^2}$

$v_f^2=347.56\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{347.56\frac{m^2}{s^2}}$

$v_f=18.64\frac{m}{s}$

Now that we have the final directional velocities, we can use the tangent function to find the angle with which the rock impacts the ground.

$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$

$\tan({\theta})=\frac{v_y}{v_x}$ .

$\theta=\tan^{-1}(\frac{v_y}{v_x})$

$\theta=\tan^{-1}(\frac{18.64}{29.13})$

$\theta=\tan^{-1}(0.64)$

$\theta=32.62^{\circ}$

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

What is the initial horizontal velocity?

Answer

The question gives the total initial velocity, but we will need to find the horizontal component using trigonometry.

To find the horizontal velocity we use the equation $v_{i}\cos(\Theta)=v_{x}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}\cos(35^{\circ})=v_{x}$

$30\frac{m}{s}*(0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

What is the initial vertical velocity?

Answer

The question gives the total initial velocity, but we will need to find the vertical component.

To find the vertical velocity we use the equation $v_{i}* \sin(\Theta)=v_{iy}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

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Question

A ball rolls off of a table with an initial horizontal velocity of $19\frac{m}{s}$ . If the table is high, how long will it take to hit the ground?

$\small g=-9.8\frac{m}{s^2}$

Answer

The problem gives us the initial horizontal velocity. This velocity will only affect the distance the ball travels in the horizontal direction; it will have no effect on the time the ball is in the air. Since there is no angle of trajectory, the ball has no initial vertical velocity.

We know the height of the table, the initial velocity, and gravity. Using these values with the appropriate motion equation, we can solve for the time.

The best equation to use is:

$\Delta y =v_it+\frac{1}{2}at^2$

We can use our values to solve for the time. Keep in mind that the displacement will be negative because the ball is traveling in the downward direction!

$-3m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-3m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-3m=(-4.9\frac{m}{s^2})t^2$

$\frac{-3m}{-4.9\frac{m}{s^2}}=t^2$

$\sqrt{0.612s^2}=\sqrt{t^2}$

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Question

A ball rolls off of a table with an initial horizontal velocity of $19\frac{m}{s}$ . If the table is high, how far from the table will it land?

$\small g=-9.8\frac{m}{s^2}$

Answer

We can solve for the horizontal distance using only the horizontal velocity: $\Delta x =v_xt$ .

We are given the value of , but we need to find the time. Time in the air will be determined by the vertical components of the ball's motion.

We know the height of the table, the initial velocity, and gravity. Using these values with the appropriate motion equation, we can solve for the time.

The best equation to use is:

$\Delta y =v_it+\frac{1}{2}at^2$

We can use our values to solve for the time. Keep in mind that the displacement will be negative because the ball is traveling in the downward direction!

$-3m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-3m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-3m=(-4.9\frac{m}{s^2})t^2$

$\frac{-3m}{-4.9\frac{m}{s^2}}=t^2$

$\sqrt{0.612s^2}=\sqrt{t^2}$

Now we have both the time and the horizontal velocity. Use the original equation to solve for the distance.

$\Delta x =v_xt$

$\Delta x =(19\frac{m}{s})(0.782s)$

$\Delta x =14.86m$

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