Question

A ball is dropped from a certain height. If it takes 0.2 s to cross the last 6.2 m before hitting the ground, find the height from which is was dropped. (Take $g=10m{s}^{-2}$)

• A. 51.2 m
• B. 25.6 m
• C. 31 m
• D. 62 m

Answer 1

The correct option is A 51.2 m

For the last part of the journey,
$s=ut+\left(1/2\right)g{t}^2$
Here $s=6.2$ m; $u=?t=0.2sg=10$ $m/s^2$
Putting these values we get, $u=30m/s$
Now, for the upper portion, $v^2=u^2+2gh$, here, $u=initialvel=0$; $v=30m/s$,
Putting the values we get $h=45$ m.
So the total height = $(45+6.2)m=51.2m$