A cricket bowler releases the ball in two different ways
(i)giving it only horizontal velocity and
(ii)giving it horizontal velocity and a small downward velocity
The speed $v_{s}$ at the time of release is the same. Both are released at a height $H$ from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

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Solution

When ball is given horizontal velocity Let us inside horizontal velocity b the time of release.
$u_{x} = v_{x}$
During trootertil motion horizontal velocity remains unchanged.
So,
$v_{x} = u_{x} = v_{x}$
In vertical direction.
$v_{y}^{2} = u y^{2} + 2 g h$ If $u_{y} = 0$
$v_{y} = \sqrt{2 g h}$
Resultant
$v = \sqrt{u_{x}^{2} + v_{y}^{2}}$
$\sqrt{v = \sqrt{v_{x}^{2} + 2 g h}}$ .......(1)
From (1) and (2)
When ball is given horizontal velocity and small downward velocity.
in horizontal direction
$v_{x} = u_{x} = v_{s}$
and in vertical
$v y^{2} = u^{2} + 2 g h$
$v = \sqrt{v_{x}^{1} + v_{y}^{1}}$
$= \sqrt{v_{s}^{2} + u^{2} + 2 g h}$ .........(2)

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