Question

A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes the pot

0.440 to pass from the top to the bottom of this window, which
is 2.20 high.

How far is the top of the window below the windowsill from which
the flowerpot fell?

Answer #1

here,

let the velocity of pot at the top of the window be u

time taken to cross window , t = 0.44 s

height of window , h = 2.2 m

using seccond equation of motion

h = u*t + 0.5 * g * t^2

2.2 = u*0.44 + 0.5 * 9.8 * 0.44^2

u = 2.84 m/s

let the height of the windowshell from the top of window be h'

initial velocity , u' = 0

final velocity at top of window , v' = u = 2.84 m/s

using third equation of motion

v^2 - u^2 = 2 * g * h

2.84^2 - 0 = 2 * 9.8 * h'

h' = 0.41 m

**the distance of the windowshell from the top of window
is 0.41 m**

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