Question

A body starts from rest and falls vertically from a height of 19.6 meters. If g=9.8 m/sec2. Then, what is the distance traveled by the body in the last 0.1 second of its motion?

Answer 1

$h=19.6,a=9.8,u=0$

$h=\frac{g{t}^2}{2}$

$t=\sqrt{\frac{2h}{g}}$

$t=2s$

we need to find $s$ in $t=1.9\to 2$

using:$s=ut+\frac{g{t}^2}{2}$

$s_2-s_1=u(t_2-t_1)+\frac{g(t_2^2-t_1^2)}{2}$

$=0+\frac{9.8}{2}(2\times 2-1.9\times 1.9)$

$=1.911m$