A small, free-falling pebble takes 0.25 s to pass by a window 1.8m high. From what height above the window was the pebble dropped?

From the diagram below, the pebble was dropped starting from y0. The pebble underwent an accelerated motion all throughout its journey.

At the moment the pebble was dropped y = 0 m and t = 0 s

At the moment the pebble was from y1 at the top of the window time of fall > 0 s and t = t

At the moment the pebble was at y2 = 1.8 m from the top of the window time of fall = t + 0.25 s

v1^2 = 2gy1

v2^2 = 2gy2

v2^2 = 2g(y2 – y1)

v2^2 = 2g(1.8)

v2^2 = 19.6 * 1.8) = 35.28 (m/s)^2

v2 = sqrt (35.28 (m/s)^2

v2 = 5.94 m/s

Solving for the time to fall at a speed of 5.94 m/s

v2 = gt

5.94 = 9.8 t

t = 5.94 / 9.8

t = 0.606 s

Solving for the height when t = 0 and v = 0

y = 1/2 gt^2

y = 4.9 * (0.606 s)^2

y = 1.8 m

Since the height when y1 is also equal to the height of the window then the pebble started falling at a height 0f 3.6 m from the ground.

v^2 = 2gh

v^2 = 2 * 9.8 * 3.6

v^2 = 70.56 (m/s)^2

v = 8.4 m/s

Solving for the time of fall

v = gt

8.4 = 9.8 t

t1 = 8.4 / 9.8

t1 = 0.8571 s

Solving for the time to cover 1.8 m

y = 4.9 t^2

1.8 = 4.9 t^2

t2^2 = 1.8 / 4.9

t^2 = 0.36734 s^2

t2 = 0.6061

Time to cover the 1.8 m window is t1 - t2 = 0.8571 s – 0.6061 s = 0.251 s

The rectangular box represents a window where the small pebble passed.