Question

Two particles of equal mass are projected simultaneously with speeds 20$m/s$ and 10$\sqrt{3}m/s$ as shown in figure. Find the maximum height reached by the centre of mass of the particles

• A. $\frac{25}{4}m$
• B. $\frac{75}{16}m$
• C. $\frac{125}{16}m$
• D. $\frac{125}{4}m$

Answer 1

The correct option is C $\frac{125}{16}m$

Given,

Speed of one particle is $10\sqrt{3}m/s$ is projected at angle ${60}^{\circ }$ therefore $y$ component of this speed is

$10\sqrt{3}sin{60}^{\circ }=10\sqrt{3}\times \frac{\sqrt{3}}{2}=\frac{10\times 3}{2}=15m/s$

Speed of another particle is $20m/s$ is projected at angle ${30}^{\circ }$ therefore $y$ component of this velocity is

$20sin{30}^{\circ }=20\times \frac{1}{2}=10m/s$

Now

velocity of Centre of mass is

$v=\frac{m\times 15+m\times 10}{2m}v=\frac{15+10}{2}v=\frac{25}{2}$

We know that

Maximum height is

$v_{max}=\frac{v^2}{2g}$

Substitute all value in above equation then,

$h_{max}=\frac{{\frac{25}{2}}^2}{2\times 10}{h}_{max}=\frac{625}{4\times 2\times 10}{h}_{max}=\frac{125}{16}m$

Hence, correct option is $C$