Question

A flowerpot falls off a windowsill and passes the window of the story below. Ignore air resistance. It takes the pot 0.360 s to pass from the top to the bottom of this window, which is 1.90 m high. How far is the top of the window below the windowsill from which the flowerpot fell?

Answer 1

Answer:

0.6286 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

The length of the window is 1.9 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\dfrac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\dfrac{1.9-\frac{1}{2}\times 9.81\times 0.36^2}{0.36}\\\Rightarrow u=3.51197\ m/s$

This velocity will be the final velocity when falling from above

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{3.51197^2-0^2}{2\times 9.81}\\\Rightarrow s=0.6286\ m$

The top of the window below the windowsill from which the flowerpot fell is at a height of 0.6286 m