Question

A boy throws a ball vertically at a velocity of 4.8 m s−1.
(a) How long is it before he catches it again?
(b) What will be the ball’s greatest height above the point
of release?

Answer 1

$V=u+at$

$-v=u+at$

$-4.8=4.8-gt$

$9t=9.8$

$t= 1 sec$

b) $v^{2}=u^{2}+2as$

$0=(4.8)^2-2(9.8)$

$S= \frac{4.8}{2\times 9.8}$

$S= 1.175m$