Question

The ceiling of a hall is $30 \,m$ high. A ball is thrown with $60 \,ms ^{-1}$ at an angle $\theta$, so that it could reach the ceiling of the hall and come back to the ground. The angle of projection $\theta$ that the ball was projected is given by

• A. $\sin \theta=\frac{1}{\sqrt{8}}$
• B. $\sin \theta=\frac{1}{\sqrt{6}}$
• C. $\sin \theta=\frac{1}{\sqrt{3}}$
• D. None of these

Answer 1

Answer: (B)

Given, $u=60 \,ms ^{-1}$
Maximum height $H$ that the ball will achieve
$=$ Height of ceiling of the hall $=30\, m$
As, maximum height, $H=\frac{u^{2} \sin ^{2} \theta}{2 g}$
$\Rightarrow 30=\frac{(60)^{2} \sin ^{2} \theta}{2 g} $
$\Rightarrow \sin ^{2} \theta=\frac{30 \times 2 g}{60 \times 60}=\frac{10}{60} $
$\left[\because g=10\, ms ^{-2}\right] $
$\Rightarrow \sin \theta=\frac{1}{\sqrt{6}}$