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.

Given that

The window height is 2m

And the window is 7.5m from the ground

Then the total height of the window from the ground is 7.5+2=9.5m

It takes the ball 0.32sec travelled pass the window.

When the ball get to the window, it has an initial velocity (u') and when it gets to the top of the window it has a final velocity (v')

Now using the equation of free fall during this window travels

S=ut-½gt² against motion.

S=2, g=9.81, t=0.32sec

Then,

S=u't-½gt²

2=u'*0.32-½*9.81*0.32²

2=0.32u'-0.5023

2+0.5032=0.32u'

Then, 0.32u'=2.5032

u'=2.5032/0.32

u'=7.82m/s

This is the initial velocity as the ball got the the window

Now, let analyse from the window bottom to the ground which is a distance of 7.5m

Using the equation of free fall again

v²=u²-2gH

In this case the final velocity (v) is the velocity when the ball reach the bottom of the window i. e u'=7.82m/s,

While u is the original initial velocity from the throw of the ball

Then,

u'²=u²-2gH

7.82²=u²-2*9.81*7.5

61.146=u²-147.15

61.146+147.15=u²

Then, u²=208.296

So, u=√208.296

u=14.43m/s

The initial velocity of the ball form the throw is 14.43m/s