Question

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