Question

The ceiling of a hall is 40m high. For maximum horizontal distance, the angle at which the ball can be thrown with a speed of 56 $m{s}^{-1}$ without hitting ceiling of the hall is $(\text{take}g=9.8m/s^2)$

• A. ${25}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B ${30}^{\circ }$

$v^2{\sin }^2\theta -2gh=0$

$\Rightarrow v^2{\sin }^2\theta =2gH$

$\Rightarrow 56\times 56{\sin }^2\theta =2\times 9.8\times 40$

Maximum height$=H$

Range$=R$

angle of projection $=\theta$

$\frac{H}{R}=\frac{1}{4}\tan \theta$

$H=40m$

$\tan \theta =\frac{1}{\sqrt{3}}$

Maximum range $=R=160\sqrt{3}m$ at this angle ${30}^0$

Hence,

option $\left(B\right)$ is correct answer.