A ball is projected vertically upwards with an initial velocity $'u'$ goes to a maximum height $'h'$ before touching the ground. What is the value of $'h'$ ?
We know the equation of motion:
$v^2=u^2+2as$
Here $u=initial\ velocity$
$s=maximum\ height=h$
$v=final\ velocity=0$
And $a=gravitational\ acceleration$
On substituting above values in the equation $v^2=u^2+2as$ we have,
$0=u^2+2( -g)h$ [Here $-g$ indicates the deceleration]
Or $u^2=2gh$
Or $h=\frac{u^2}{2g}$
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