A cricket bowler releases the ball in two different ways
(a) giving it only horizontal velocity and
(b) 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 .

Explanation

(a) When ball is given only horizontal velocity Horizontal velocity at the time of release
`(u_(x)) = v_(s)`
During projectile motion , horizontal velocity remains unchanged .
Therefore , `" " v_(x) = u_(x) = v_(s)`
In vertical direction , `" " v_(y)^(2) = u_(y)^(2) + 2gH`
`v_(y) = sqrt(2gH ) " " ( because u_(y) = 0)`

`therefore` Resultant speed of the ball at bottom ,
`v = sqrt(v_(x)^(2) + v_(y)^(2))`
`= sqrt(v_(s)^(2) + 2gH) " " ..... (i)`
(b) When ball is given horizontal velocity and a small downward velocity

Let the ball be given a small downward velocity u .
In horizontal direction `" " v'_(x) = u_(x) = v_(s)`
In vertical direction `" " v'_(y)^(2) = u^(2) + 2gH`
or `" " v'_(y) = sqrt(u^(2) + 2gH)`
`therefore` Resultant speed of the ball at the bottom
`v' = sqrt(v'_(x)^(2) + v'_(y)^(2)) = sqrt(v_(s)^(2) + u^(2) + 2gH) " " .....(ii)`
From Eqs . (i) and (ii) , we get `v' gt v`