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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Updated on: 10-Oct-2022

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