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Question

A cricket ball is thrown at a speed of $$28 \ ms^{-1}$$ in a direction $$30^0$$ 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.


Solution

Given :   $$u = 28 \ m/s$$          $$\theta = 30^o$$

(a) :  Maximum height   $$H = \dfrac{u^2\sin^2\theta}{2g}$$
$$\implies \ H = \dfrac{28^2\times 0.5^2}{2\times 9.8} =10 \ m $$

(b) :  Time of flight   $$T = \dfrac{2u\sin \theta}{g}$$
$$\implies \ T = \dfrac{2\times 28\times sin 30}{9.8}  =  2.86 \ s $$

(c) :  Range    $$R = u\cos\theta\times T $$
$$\implies \ R = 28\times \cos30^o\times 2.86 = 69.27 \ m$$

Physics

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