An object is thrown vertically upwards and rises to a height of $10cm$. Calculate
(i) the velocity with which the object was thrown upwards and
(ii) the time taken by the object to reach the highest point.

Given,

Height $h=10cm=0.1m$

At maximum height, final velocity $v=0$

Apply second kinematic equation

$v^2-u^2=2ah$

$0-u^2=2(-g)\times h$

$u=\sqrt{2gh}=\sqrt{2\times 10\times 0.1}=\sqrt{2}m/s$

Apply first kinematic equation

$v=u+at$

$0=u-gt$

$t=\frac{u}{g}=\frac{\sqrt{2}}{10}\cong 0.14\sec$

Initial velocity is $\sqrt{2}m/s$ and time required tor each maximum height is $0.14\sec$.