Question

Two particles are projected from the ground simultaneously with speeds $20m/s$ and $\frac{20}{\sqrt{3}}m/s$at angles ${30}^{\circ }$ and ${60}^{\circ }$ with the horizontal in the same direction. The maximum distance between them till both of them strike the ground is approximately ($g$ = $10m/s^2$)

• A. 23.1 m
• B. 16.4 m
• C. 30.2 m
• D. 10.4 m

Answer 1

The correct option is A

23.1 m

Component of velocities of both the particles in vertical direction are equal. Therefore, their time of flights are equal and their relative motion is in horizontal direction only. Thus, the maximum distance between them is equal to the differencein their horizontal ranges.

$R_1$ = $\frac{(20{)}^{2}sin{60}^{\circ }}{g}$ = $\frac{\left(400\right)\sqrt{3}}{10\times 2}$ = $34.64m$

$R_2$ = $\frac{{\left(\frac{20}{\sqrt{3}}\right)}^{2}sin{120}^{\circ }}{g}$ = $\frac{400\times \sqrt{3}}{10\times 3\times 2}$ = $11.55m$

$\therefore$ $S_{max}$ = $34.64-11.55$=23.1 m