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Question: The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may...

The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 m s-1 without hitting the ceiling of the hall is

A

25025^{0}

B

30o30^{o}

C

45o45^{o}

D

60o60^{o}

Answer

30o30^{o}

Explanation

Solution

Here, u=56ms−1u = 56ms^{- 1}

Let θ\thetabe the angle of projections with the horizontal to have maximum range, with maximum height = 40 m

Maximum height, H=u2sin⁡2θ2gH = \frac{u^{2}\sin^{2}\theta}{2g}

40=(56)2sin⁡2θ2×9.840 = \frac{(56)^{2}\sin^{2}\theta}{2 \times 9.8}

sin⁡2θ=2×9.8×40(56)2=14orSinθ=12\sin^{2}\theta = \frac{2 \times 9.8 \times 40}{(56)^{2}} = \frac{1}{4}orSin\theta = \frac{1}{2}

θ=sin⁡−1(12)=30∘\theta = \sin^{- 1}\left( \frac{1}{2} \right) = 30{^\circ}