Question

A particle is projected with a velocity $u$ at two different angles which are complementary to each other. If $H_1$ and $H_2$ are the heights attained by them in each case, find the range of the particle.

• A. $4\sqrt{H_1{H}_2}$
• B. $2\sqrt{H_1{H}_2}$
• C. $\sqrt{H_1{H}_2}$
• D. $\frac{1}{2}\sqrt{H_1{H}_2}$

Answer 1

The correct option is A $4\sqrt{H_1{H}_2}$

Let $\theta$ and $90-\theta$ be the two complimentary angles of projections,
When angle of projection is $\theta$:
Maximum height of a projectile, $H_1=\frac{u^2{\sin }^2\theta }{2g}$
Range of projectile, $R=\frac{2u^2\sin \theta \cos \theta }{g}$
When angle of projection is $90-\theta$:
Maximum height of a projectile, $H_2=\frac{u^2{\sin }^2(90-\theta )}{2g}=\frac{u^2{\cos }^2\theta }{2g}$
Range of a projectile, $R=\frac{2u^2\sin (90-\theta )\cos (90-\theta )}{g}$
$=\frac{2u^2\sin \theta \cos \theta }{g}$
$\therefore$ Range is same for both the angles of projection
$R^2=\frac{4u^2{\sin }^2\theta {\cos }^2\theta }{g^2}\Rightarrow R^2=16\left(\frac{u^2{\sin }^2\theta }{2g}\right)\left(\frac{u^2{\cos }^2\theta }{2g}\right)\Rightarrow R^2=16H_1{H}_2\therefore R=4\sqrt{H_1{H}_2}$