Question

# A cricket ball is thrown at a speed of $$28 \ ms^{-1}$$ in a direction $$30^0$$ above horizontal. Calculatea) The maximum heightb) 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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