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Question

A cricket ball is thrown at a speed of 28 ms1 in a direction 300 above horizontal. Calculate
a) The maximum height
b) The time taken by the ball to return to the same level.
c) The distance from the thrower to the point where ball returns to the same level.

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Solution

Given : u=28 m/s θ=30o

(a) : Maximum height H=u2sin2θ2g
H=282×0.522×9.8=10 m.

(b) : Time of flight T=2usinθg
T=2×28×sin309.8=2×28×0.59.8=2.86 s.

(c) : Range R=ucosθ×T
R=28×cos30o×2.86=28×32×2.86=69.27 m.

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