1. A cricket ball is hit vertically and returns to the ground 6 s later. Calculate: (i) Maximum
height reached by the ball. (ii) Initial velocity of the ball.
Answers
Explanation:
A cricket ball hits vertically upwards and returns to the ground 6 seconds later. What is the max height and initial velocity?
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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.