A ball is dropped from somewhere above a window that is 2.00 m in height. As it falls, it is visible to a person boxing through the window for 200 ms as it passes by the 2.00 m height of the window. From what height above the top of the window was the ball dropped?

4.14 m

Explanation:

In the last leg of the journey the ball covers 2 m in 2ms or 0.2 s.

Let in this last leg, u be the initial velocity.

s = ut + 1/2 g t²

2 =.2 u +.5 x 9.8 x. 04

u = 9.02 m / s.

Let v be the final velocity in this leg

v² = u² + 2 g s

v² = (9.02) ² + 2 x 9.8 x 2

= 81.36 + 39.2

v = 10.97 m / s

Now consider the whole height from where the ball dropped. Let it be h.

Initial velocity u = 0

v² = u² + 2gh

(10.97) ² = 2 x 9.8 h

h = 6.14 m

Height from window

= 6.14 - 2m

= 4.14 m