A falling stone takes delta t = 0.32s to travel past a window 2.2m Tall. From what height above the top of the window did the stone fall?

The height above the top of the window is 1.44 m

Explanation:

Given;

time of motion, t = 0.32 s

height traveled at the given time, h = 2.2m

determine the initial velocity of the stone;

h = ut + ¹/₂gt²

2.2 = u(0.32) + ¹/₂ x 9.8 x 0.32²

2.2 = 0.32u + 0.502

0.32u = 2.2 - 0.502

0.32u = 1.698

u = 1.698 / 0.32

u = 5.31 m/s

This initial velocity on top of the window becomes the final velocity from the height above the window.

v² = u² + 2gh

where;

u is the initial velocity of the stone from the height above the window;

5.31² = 0 + (2 x 9.8)h

19.6h = 28.196

h = 28.196/19.6

h = 1.44 m

Therefore, the height above the top of the window is 1.44 m

where are the vectors ?

a. are