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 horizontal velocity = `upsilon_(s)`
and there is no downward velocity given ,
As shown in `upsilon_(z) = sqrt(2 g H)`
`:.` velocity on hitting the ground
`upsilon sqrt(upsilon_(s)^(2) + upsilon_(z)^(2)) = sqrt(upsilon_(s)^(2) + g H`
(b) When horizontal velocity = `upsilon_(s)`
and downward velocity = u
as shown
Then `u_(z) = sqrt(u^(2) + 2 gH`
velocity on hitting the ground `upsilon = sqrt(upsilon_(s)^(2)+upsilon_(z)^(2)) = sqrt(upsilon_(s)^(2) + u^(2) + 2 gh)` Clearly `upsilon' gt upsilon`
Note If u is taken as negligibly small then `upsilon = upsilon`
.