Question

A flowerpot falls off a windowsill and passes the win- dow of the story below. Ignore air resistance. It takes the pot 0.380 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 window- sill from which the flowerpot fell?

Answer 1

Answer:

d = 0.50 m

Explanation:

Let say the speed at the top and bottom of the window is

$v_1 \: and \: v_2$ respectively

now we have

$d = \frac{v_1 + v_2}{2}t$

$1.90 = \frac{v_1 + v_2}{2} (0.380)$

$v_1 + v_2 = 10$

also we know that

$v_2 - v_1 = 9.8(0.380)$

$v_2 - v_1 = 3.72$

now we have from above equations

$v_2 = 6.86 m/s$

$v_1 = 3.14 m/s$

now the distance from which it fall down is given as

$v_f^2 - v_i^2 = 2ad$

$3.14^2 - 0^2 = 2(9.8)d$

$d = 0.50 m$