Question

Two stones are projected so as to reach the same distance from the point of projection on a horizontal surface. The maximum height reached by one exceeds the other by an amount equal to half the sum of the heights attained by them. Then the angles of projection for the stones are (in degrees)

• A. $30,60$
• B. $0,90$
• C. $70,20$
• D. $80,10$

Answer 1

The correct option is A $30,60$

Let the height attained by the two stones be $h_1\&h_2$ respectively.

By given condition,

$h_1-h_2=\frac{1}{2}(h_1+h_2)$

$2h_1-2h_2=h_1+h_2$

$h_1={3h}_2$ (Eq-1)

But for maximum height

$h=\frac{u^2{\sin }^2\theta }{2g}$

But,it is also given that their ranges are equal.

$R_1=R_2$

$R=\frac{u^2\sin 2\theta }{g}$

$\Rightarrow \frac{u_1^2\sin 2{\theta }_1}{g}=\frac{u_2^2\sin 2{\theta }_2}{g}$

$\Rightarrow \frac{u_2^2}{u_1^2}=\frac{\sin 2{\theta }_1}{\sin 2{\theta }_2}$ (Eq-2)

Also from Eq (1)

$u_1^2{\sin }^2{\theta }_1=3u_2^2{\sin }^2{\theta }_2$

${\sin }^2{\theta }_1=\frac{3u_2^2}{u_1^2}{\sin }^2{\theta }_2$

From Eq-(2),

$\frac{\sin 2{\theta }_1}{\sin 2{\theta }_2}=\frac{u_2^2}{u_1^2}$

$\therefore {\sin }^2{\theta }_1=\frac{3\times \sin 2{\theta }_1\times {\sin }^2{\theta }_2}{{\sin }^2{\theta }_2}$

$\therefore {\sin }^2{\theta }_1=\frac{3\times 2\sin {\theta }_1.\cos {\theta }_1\times {\sin }^2{\theta }_2}{2\sin {\theta }_2\cos {\theta }_2}$

$\therefore \frac{\sin {\theta }_1}{\cos {\theta }_1}=\frac{3\sin {\theta }_2}{\cos {\theta }_2}$

$\therefore \tan {\theta }_1=3\tan {\theta }_2$

$\frac{\tan {\theta }_1}{\tan {\theta }_2}=3$

$\Rightarrow \frac{\tan {\theta }_1}{\tan {\theta }_2}=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}$

$\therefore {\theta }_1={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)={30}^o$

${\theta }_2={\tan }^{-1}\left(\sqrt{3}\right)={60}^o$