College Physics by Openstax Chapter 2 Problem 54

A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past the window. What was the ball’s initial velocity? Hint: First consider only the distance along the window, and solve for the ball’s velocity at the bottom of the window. Next, consider only the distance from the ground to the bottom of the window, and solve for the initial velocity using the velocity at the bottom of the window as the final velocity.

Solution:

First, we have the position 1 where the motion starts. Here, we know that y<sub>1</sub>=0, t<sub>1</sub>=0, and v<sub>y1</sub>=0. Position 2 is at the bottom of the window. We know that it is 7.50 meters from where the motion started. So we have y<sub>2</sub>=7.50 meters. We do not know the time and velocity at this point. Then we have position 3 at the top of the window where the overall height is 9.50 meters, y<sub>3</sub>=9.50. We also do not know the velocity and time elapsed in this position.
Figure A

Consider Figure A. We shall be considering the three positions shown.

First, we have position 1 where the motion starts. Here, we know that y1=0 and t1=0, but we do not know vy1.

Position 2 is at the bottom of the window. We know that it is 7.50 meters from where the motion started. So we have y2=7.50 meters. We do not know the time and velocity at this point.

Then we have position 3 at the top of the window where the overall height is 9.50 meters, y3=9.50. We also do not know the velocity and time elapsed in this position.

Consider positions 2 and 3. The initial position in this case is at position 2 and the final position is at position 3. We know that the difference of time between this two positions is 0.312 seconds. We can say that

t_3 =t_2+0.312 \ \text{s} \\
t_3-t_2 = 0.312\ \text{s}

Using the same 2 positions still, we have

\begin{align*}

y_3 & = y_2 + v_{y_2} \Delta t+\frac{1}{2}a\left(  \Delta t \right)^2 \\
9.50\ \text{m} & = 7.50\ \text{m} +  v_{y_2} \left( t_3-t_2 \right)+\frac{1}{2}a\left( t_3-t_2 \right)^2 \\
9.50\ \text{m}-7.50\ \text{m} & = v_{y_2}\left( 0.312\ \text{s} \right)+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 0.312\ \text{s} \right)^2\\
2.00\ \text{m} & = 0.312\ \text{s} \left( v_{y_2} \right)-0.4775\ \text{m} \\
 0.312\ \text{s} \left( v_{y_2} \right) & = 2.00\ \text{m}+0.4775\ \text{m} \\
 0.312\ \text{s} \left( v_{y_2} \right) & = 2.4775\ \text{m} \\
v_{y_2}& =\frac{2.4775\ \text{m}}{0.312\ \text{s}} \\
v_{y_2}& = 7.94\ \text{m/s}

\end{align*}

We have computed the velocity of the ball at the bottom of the window.

Next, we shall consider positions 1 and 2. In this consideration, position 1 will be considered the initial position while position2 is the final position. 

\begin{align*}

\left( v_{y_2} \right)^2  &  = \left( v_{y_1} \right)^2 +2a \Delta y \\
\left( 7.94\ \text{m/s} \right)^2 & = \left( v_{y_1} \right)^2 + 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)\\
\left( v_{y_1} \right)^2 & = \left( 7.94\ \text{m/s} \right)^2- 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)\\
v_{y_1} & = + \sqrt{ \left( 7.94\ \text{m/s} \right)^2- 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)} \\
v_{y_1} & = + 14.5\ \text{m/s}\qquad {\color{DarkOrange} \left( \text{Answer} \right) }

\end{align*}

\therefore The ball’s initial velocity is about 14.5 m/s upward.

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