Question

A cricket ball is hit vertically
upwards and returns to ground
6 s later. Calculate (i) maximum
height reached by the ball.
(ii) initial velocity of the ball.​

Answer 1

Answer:

If air resistance is neglected, the time of rise of a projectile is equal to its time of fall. Since the ball returns to ground after 6 s, the time of fall is 6s/2 or 3 s. The maximum height is solved using dy = 1/2gt^2 where dy is the maximum height and g = 9.8 m/s^2.

Solving for dy

dy = 1/2gt^2

dy = 1/2 * 9.8 m/s^2 * (3s)^2

dy = 4.9 * 9

dy = 44.1 m

Solving for the initial velocity Vo

dy = (Vo sin(theta)^2 / 2g

dy = Vo^2 * (sin 90 degrees)^2 / 19.6 m/s^2

44.1 = Vo^2 * 1 / 19.6

Vo^2 = 44.1 * 19.6

Vo^2 = 864.36

Vo = sqrt (864.36)

Vo = 29.4 m/s

The max height = 44.1 m and the initial velocity Vo = 29.4 m/s.

Explanation:

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Answer 2

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