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

d = 0.50 m

Step-by-step explanation:

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

`v_1 \: and \: v_2` respectively

now we have

`d = (v_1 + v_2)/(2)t`

`1.90 = (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`