$A$ 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.

(NONE) Let the initial speed of release be $V_s$ in both cases. We can use the principle of conservation of mechanical energy.
Total mechanical energy at the point of release = Total mechanical energy at the point of impact.
Let the ground be the reference level for potential energy $(PE = 0)$.
Initial energy $E_i = \frac{1}{2} m V_s^2 + mgH$.
Final energy $E_f = \frac{1}{2} m v^2 + 0$,where $v$ is the final speed.
Since energy is conserved,$E_i = E_f$.
$\frac{1}{2} m V_s^2 + mgH = \frac{1}{2} m v^2$.
$v^2 = V_s^2 + 2gH$.
$v = \sqrt{V_s^2 + 2gH}$.
Since the final speed $v$ depends only on the initial speed $V_s$,the height $H$,and the acceleration due to gravity $g$,and all these values are identical in both cases $(a)$ and $(b)$,the final speed when the ball hits the ground will be the same in both cases.