Question

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 ms^-1without hitting the ceiling of the hall is

Moderate

Solution

$h = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{(56)^{2} \sin^{2} θ}{19 .6} \begin{array}{l} \sin^{2} θ = \frac{40 \times 19 .6}{(56)^{2}} = \frac{1}{4} \\ \sin θ = \frac{1}{2} or θ = 30^{\circ} \end{array}$

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