Question

Two particles projected from the same point with same speed $u$ at angles of projection $\alpha$ and $\beta$ strike the horizontal ground at the same point. If $h_1$ and $h_2$ are the maximum heights attained by the projectile, $R$ is the range for both $t_1$ and $t_2$ are their times of flights, respectively, then

• A. $\alpha +\beta =\frac{\pi }{2}$
• B. $R=4\sqrt{h_1{h}_2}$
• C. $\frac{t_1}{t_2}=\tan \alpha$
• D. $\tan \alpha =\sqrt{\frac{h_1}{h_2}}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $R=4\sqrt{h_1{h}_2}$
B $\tan \alpha =\sqrt{\frac{h_1}{h_2}}$
C $\frac{t_1}{t_2}=\tan \alpha$
D $\alpha +\beta =\frac{\pi }{2}$

From above concept $\alpha +\beta =\frac{\pi }{2}$

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

$h_1=\frac{u^2{\sin }^2\alpha }{2g}$

$h_2=\frac{u^2{\cos }^2\alpha }{2g}$

multiplying $h_1$ and $h_2$

$h_1{h}_2=\frac{u^2{\sin }^2\alpha u^2{\cos }^2\alpha }{g^2}$

$h_1{h}_2=\frac{u^4{\sin }^{2}2\alpha }{4g^2}$

$4h_1{h}_2=(\frac{R}{2}{)}^2$

$R=4\sqrt{h_1{h}_2}$

$t_1=\frac{2u\sin \alpha }{g}$

$t_2=\frac{2u\cos \alpha }{g}$

$\frac{t_1}{t_2}=\tan \alpha$

$\frac{h_1}{h_2}={\tan }^2\alpha$

$\tan \alpha =\sqrt{\frac{h_1}{h_2}}$