Example 4.9 A cricket ball is thrown at a speed of 28 ms in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level.
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Given ,
u = 28 m/s ,. θ = 30°
(a) Maximum height
H = (u^2 sin^2θ)/ 2g
=> H = (28^2 . sin^2 30°) / (2×9.8)
=> H = (28^2 . 1/4) / (2×9.8)
=> H = 10 m
(B) Time of flight
T = (2 u sinθ)/g
=> T = (2 × 28 × sin 30°) / 9.8
=> T = 2.86 s
(C) Range
R = u cosθ T
=> R = 28 × cos30° × 2.86
=> R = 69.27 m
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