Question

You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone video taken in a 4th-floor apartment, which happened to capture the flowerpot as it fell past a window. In a span of 8 frames (captured at 30 frames per second), the flowerpot falls 0.84 of the height of the window. You visit the apartment and measure the window to be 1.27 m tall.

Required:
Assume the flowerpot was dropped from rest. How high above the window was the flowerpot when it was dropped?

Answer 1

0.37 m

Step-by-step explanation:

Given :

Window height,
`h_1` = 1.27 m

The flowerpot falls 0.84 m off the window height, i.e.

`h_2` = (1.27 x 0.84 ) m in a time span of
`$t=(8)/(30)$` seconds.

Assuming that the speed of the pot just above the window is v then,

`h_2=ut+(1)/(2)gt^2`

`$(1.27 * 0.84) = v * \left( (8)/(30) \right) + (1)/(2) * 9.81 * \left( (8)/(30) \right)^2$`

`$v=\left((30)/(8)\right) \left[ (1.27 * 0.84) - \left( (1)/(2) * 9.81 * \left( (8)/(30 \right)^2 \right) \right])$`

`$v= 2.69$` m/s

Initially the pot was dropped from rest. So, u = 0.

If it has fallen from a height of h above the window then,

`$h = (v^2)/(2g)$`

`$h = ((2.69)^2)/(2 * 9.81)$`

h = 0.37 m