Answer to Question #93486 in Mechanics | Relativity for E

Question

A baseball is dropped from the roof of a building. After some time, it flies past a window of height 2.20m. if the time it takes the baseball to fall past the window is 1/3s, how far above the window is the roof?

Answer 1

Let $v$ be the velocity of ball just before it comes to the upper part of the window

Using 2nd equation of motion when the ball passing through window

$s=(u\times t)+(\frac{a}2\times t^2)$

now

$s=2.20m$

t$=\frac{1}{3}sec$

$a=g=9.8\frac{m}{{sec}^2}$

$2.20=(v\times{\frac1 {3}})+(\frac{9.8}{2}\times\frac{1}{3}\times\frac{1}{3})$

hence, $v=4.95\frac{m}{sec}$

Using 3rd equation of motion from starting point to window

$v^2-u^2=2\times a\times s$

$u=0$ as ball started from rest

$v=4.95\frac{m}{sec}$

$a=g=9.8\frac{m}{{sec}^2}$

so,

$4.95^2=2\times 9.8\times s$

$s=1.25m$

Roof is $1.25m$  above the window