Question

Two particles A and B are projected from the same point in different directions in such a manner that vertical components of their initial velocities are same in above figure. Find the the ratio of range.

Answer 1

We know that the range of projectile,

$R=\frac{u^2\sin 2\theta }{g}=\frac{\left(2u\sin \theta \right)}{g}\times u\cos \theta =\frac{2u_x{u}_y}{g}$

As, $(u_y{)}_1=(u_y{)}_2\Rightarrow u_A\sin {\theta }_A=u_B\sin {\theta }_B$

or $\frac{u_A}{u_B}=\frac{\sin {\theta }_B}{\sin {\theta }_A}$

$\frac{R_A}{R_B}=\frac{2\left(u_A\sin {\theta }_A\right)\left(u_A\cos {\theta }_A\right)/g}{2\left(u_{B}sin{\theta }_B\right)\left(u_{B}cos{\theta }_B\right)/g}$

$=\left[\frac{u_A\cos {\theta }_A}{u_B\cos {\theta }_B}\right]$

$=\frac{\sin {\theta }_B\cos {\theta }_A}{\sin {\theta }_A\cos {\theta }_B}=\frac{\tan {\theta }_B}{\tan {\theta }_A}$