Question

A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance It takes the pot 0.420 s to pass from the top to the bottom of this window, which is 1.90 m high.1. How far is the top of the window below the windowsill from which the flowerpot fell?

Answer 1

Answer:

the distance of the top of the window below the windowsill from which the flowerpot fell is 0.31 m

Explanation:

given information:

time, t = 0.420 s

height, h = 1.9 m

the the vertical motion equation:

h = $v_{0}$t + $\frac{1}{2} gt^{2}$

where

h = height (m)

$v_{0}$ = initial velocity (m/s)

t = time (s)

g = gravitational force (9.8 $m/s^{2}$)

first we calculate the velocity of the flowers pot when it reaches top window

h = $v_{0}$t + $\frac{1}{2} gt^{2}$

1.9 = $v_{0}$(0.42) + ($\frac{1}{2} 9.8 0.42^{2}$)

$v_{0}$ = (1.9 - ($\frac{1}{2} 9.8 0.42^{2}$))/0.42

   = 2.47 m/s

now we can find the distance of the windowsil

$v^{2} =v_{0} ^{2}$ + 2gh

v = 0, thus

2gh = - $v_{0}^{2}$

h = -$v_{0}^{2}$/2g

  = $(- 2.47^{2} )$ / (2 x 9.8)

  = 0.31 m