Question

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

Answer 1

Initial velocity, $u=50m{s}^{-1}$

At maximum height final velocity, $v=0$

Apply first kinematic equation of motion

$v=u+at$

$0=50-10t$

$t=5\sec$

Apply second kinematic equation of motion

$v^2-u^2=2as$

$0^2-{50}^2=2(-10)h$

$h=125m$

Hence, at maximum height final velocity is zero, time taken $5\sec$and maximum height is $125m$