Solveeit Logo

Question

Physics Question on projectile motion

The ceiling of a hall is 40m40\,m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56ms156\,ms^{-1} without hitting the ceiling of the hall is

A

25?25^?

B

30?30^?

C

45?45^?

D

60?60^?

Answer

30?30^?

Explanation

Solution

Here, u=56ms1u = 56\,ms^{-1} Let θ\theta be the angle of projection with the horizontal to have maximum range, with maximum height =40m= 40\,m Maximum height, H=u2sin2θ2gH=\frac{u^{2}\,sin^{2}\,\theta}{2g} 40=(56)2sin2θ2×9.840=\frac{\left(56\right)^{2}\,sin^{2}\,\theta}{2\times9.8} sin2θ=2×9.8×40(56)2=14sin^{2}\,\theta =\frac{2\times9.8\times40}{\left(56\right)^{2}}=\frac{1}{4} or sinθ=12sin\,\theta =\frac{1}{2} or θ=sin1(12)\theta=sin^{-1}\left(\frac{1}{2}\right) =30?=30^{?}