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 projectiles, $R$ be the range for both and $t_1$ and $t_2$ be their time 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 $\alpha +\beta =\frac{\pi }{2}$
B $\frac{t_1}{t_2}=tan\alpha$
C $R=4\sqrt{h_1{h}_2}$
D $tan\alpha =\sqrt{\frac{h_1}{h_2}}$

$A$ $B$
$\theta =\alpha \theta =\beta$
$R=RR=R$
$H=h_{1}H=h_2$
$u=uu=u$
$t_1=\frac{2u\sin \alpha }{g}{t}_2=\frac{2u\sin \beta }{g}$
$\frac{t_1}{t_2}=\frac{2u\sin \alpha }{g}\cdot \frac{g}{2u\sin \beta }=\frac{\sin \alpha }{\sin (\pi /2-\alpha )}=\tan \alpha$ (Option C)
$R_1=\frac{u^2\sin 2\alpha }{g}{R}_2=\frac{u^2\sin 2\beta }{g}$
$\because R_1=R_2\Rightarrow \sin 2\alpha =\sin 2\beta$
$\Rightarrow 2\alpha =\pi -2\beta$
$\Rightarrow \alpha +\beta =\pi /2$ (Option A)
$H_1=\frac{u^2{\sin }^2\alpha }{2g}{H}_2=\frac{u^2{\sin }^2\beta }{2g}$
$\sqrt{\frac{H_1}{H_2}}=(\frac{u^2{\sin }^2\alpha }{2g}\cdot \frac{2g}{u^2{\sin }^2\beta }{)}^{1/2}$
$\sqrt{\frac{H_1}{H_2}}=\frac{\sin \alpha }{\sin \beta }=\frac{\sin \alpha }{\sin (\pi /2-\alpha )}=\tan \alpha$ (Option D)
$\sqrt{H_1{H}_2}=\frac{u^2\sin \alpha \sin \beta }{2g}=\frac{2u^2}{4g}\sin \alpha \sin (\pi /2-\alpha )$
$\sqrt{H_1{H}_2}=\frac{u^2}{4g}\sin 2\alpha$
$4\sqrt{H_1{H}_2}=R$ (Option B)