# Falling Objects| Kinematics| Ball Passing a High Window| College Physics| Problem 2.54

### 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 1.30 s to go past the window. What was the ball’s initial velocity?

Solution:

The time it takes the ball to reach the top of the window, at a height of 9.5 m, will be the amount of time it takes to reach the bottom of the window, plus the time to traverse the window, 1.30 s. So let’s denote the time to the top of the window as $t_{top}$, and the time to the bottom of the window $t_{B}$. We then have $t_{top}=t_B+1.30$. We will call the height to the bottom of the window $y_1$, and the height to the top of  the window $y_2$

Then we have $y_2=y_{2,0}+v_ot+\frac{1}{2}at^2$ $9.50=0+v_o\left(t_{top}\right)+\frac{1}{2}\left(-9.80\right)\left(t_{top}\right)^2$

We also have an equation for $y_1$, $y_1=v_0t_B+\frac{1}{2}a\left(t_B\right)^2$ $7.50=v_0t_B-4.90\left(t_B\right)^2$

So, now we have three equations in 3 unknowns $\left(t_B,\:t_{top},\:and\:v_0\right)$. Solving, we get the value of $v_o=14.5\:m/s$

The ball’s initial velocity is 14.5 m/s.