Question

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.

Answer 1

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.

$v_y^2=u{y}^2+2gh$ If $u_y=0$

$v_y=\sqrt{2gh}$

Resultant

$v=\sqrt{u_x^2+v_y^2}$

$\sqrt{v=\sqrt{v_x^2+2gh}}$.......(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

$v{y}^2=u^2+2gh$

$v=\sqrt{v_x^1+v_y^1}$

$=\sqrt{v_s^2+u^2+2gh}$.........(2)