Question

$A$ and $B$ shows path of two projectile, then:

• A. $\tan \theta =\sqrt{\frac{H_1}{H_2}}$
• B. $\tan \theta =\sqrt{\frac{H_2}{H_1}}$
• C. $H_1^2+H_2^2=R^2$
• D. $R=\frac{H_1^2}{H_2}$

Answer 1

The correct option is A $\tan \theta =\sqrt{\frac{H_1}{H_2}}$

According to question

Range for both projectile is same.

For $A$

$r=\left(u\cos \theta \right).t$

$H_1=u\sin \theta .t-\frac{1}{2}g{t}^2$

where $t=$ time to travel, $H=$ Height.

$\because$ at $H$ height, $V=0$

$\therefore u\sin \theta -gt=0$

$t=\frac{u\sin \theta }{g}$

$=u\sin \theta \left(\frac{u\sin \theta }{g}\right)-\frac{1}{2}g\times {\left(\frac{u\sin \theta }{g}\right)}^2$

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

$H_1=\frac{u^2{\sin }^2\theta }{2g}$

$R=u\cos \theta \left(\frac{2u\sin \theta }{g}\right)$ $[$ Because the time will double the time to reach height $H_1]$

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

$=\frac{R}{H_1}=4\cot \theta$

For $B$

$R=u\cos \alpha .t$ $[$Let the angle of projection for $B$ be $\alpha ]$

$\frac{R}{H}=4\cot \alpha$ $[$ Similarly, as $A]$

$\because$ Range is same.

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

$\sin 2\theta =\sin (\pi -2\alpha )$

$\theta +\alpha =\frac{\pi }{2}$

$\alpha =\frac{\pi }{2}-\theta$

$\therefore \frac{R}{H_2}=4\cot \alpha$

$\frac{R}{H_2}=4\tan \theta$$⟶\left(2\right)$

$\therefore \frac{4\tan \theta H_2}{4\cot \theta H_1}=1$ $[$ From equation $\left(1\right)$ and $\left(2\right)]$

$\tan \theta =\sqrt{\frac{H_1}{H_2}}$