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

Answer:

0.37 m

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 $t=\frac{8}{30}$   seconds.

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

$h_2=ut+\frac{1}{2}gt^2$

$(1.27 \times 0.84) = v \times \left( \frac{8}{30} \right) + \frac{1}{2} \times 9.81 \times \left( \frac{8}{30} \right)^2$

$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$

$v= 2.69$ 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,

$h = \frac{v^2}{2g}$

$h = \frac{(2.69)^2}{2 \times 9.81}$

h = 0.37 m