Question

A ball is projected vertically upward with a speed of $50m/s$. Find (a) the maximum height,
(b) the time to reach the maximum height
(c) the speed at half the maximum height. Take $g=10m/s^2$.

Answer 1

$u=50m/s$
(a) At the top, the final velocity $=0$
$v^2=u^2+2as$
$a=-g=-10m/s^2$
$0=(50{)}^2+2S\ast 10$
$\frac{+2500}{+20}=S$
$S=125m$
(b) Time$=$?
Using $v=u+at$
$0=50+(-10)\times t$
$t=\frac{+50}{+10}$
$t=5\sec$
(c) Half of max height $=\frac{125}{2}m$
$v^2=u^2+2as$
$v^2=(50{)}^2+2\times (-10)\times \frac{125}{2}$
$v^2=2500-1250$
$v=\sqrt{1250}$