Question

A boy standing on a building of height $20\text{m}$ throws a stone with a velocity of $15{\text{ms}}^{-1}$. What should be the angle of projection so that the stone has a range of $30\text{m}$ ?
(Take $g=10m/s^2$)

• A. ${\tan }^{-1}\frac{3}{2}$
• B. ${\tan }^{-1}\frac{3}{4}$
• C. ${30}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is A ${\tan }^{-1}\frac{3}{2}$

Let $\theta$ be the angle of projection of the stone from the building.
Initial velocity of stone $=15{\text{ms}}^{-1}$
Initial vertical component of velocity $=15\sin \theta {\text{ms}}^{-1}$
Initial horizontal component of velocity $=15\cos \theta {\text{ms}}^{-1}$
Displacement of the stone in vertical direction when it reaches the ground $=-20\text{m}$
(Taking downward direction as positive).
Using second equation of motion, we know
$h=u_{y}t+\frac{1}{2}a{t}^2$
$\Rightarrow -20=15\sin \theta t-\frac{1}{2}g{t}^2$
$\Rightarrow -20=15\sin \theta t-5t^2$
$\Rightarrow -4=3\sin \theta t-t^2\text{(i)}$
Range $=R=30\text{m}$
We know that,
$R=u_{x}t=15\cos \theta \times t$
$\Rightarrow 30=15\cos \theta \times t$
$\Rightarrow t=\frac{2}{\cos \theta }\text{(ii)}$
Using (ii) in (i), we get
$-4=6\tan \theta -4{\sec }^2\theta$
By using the identity, ${\sec }^2\theta =1+{\tan }^2\theta$
$-4=6\tan \theta -4-4{\tan }^2\theta$
$\Rightarrow 0=4\tan \theta (\frac{3}{2}-\tan \theta )$
$\Rightarrow \tan \theta =0$ or $\tan \theta =\frac{3}{2}$
Angle of projection can’t be $0^{\circ }$.
$\therefore \theta ={\tan }^{-1}\frac{3}{2}$
Thus, the correct answer is option (a)