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Question
A cricket bowler releases the ball in two different ways
- giving it only horizontal velocity, and
- giving it horizontal velocity and a small downward velocity. The speed vs 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.
Solution
a. When a ball is given only horizontal velocity at the time of release (ux) = vs
During projectile motion. horizontal velocity remains unchanged,
Therefore, vx = ux = vs
In vertical direction, `v_y^2 = u_y^2 + 2gH`
`v_y = sqrt(2gH)` .....(∵ uy = 0)
∴ Resultant speed of the ball at the bottom,
`v = sqrt(v_x^2 + v_y^2)`
= `sqrt(v_s^2 + 2gH)` ......(i)
b. When the 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)`
∴ 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 equations (i) and (ii), we get v' > v
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