Question

A particle is projected with speed $10m/s$ at an angle ${60}^o$ with the horizontal. Then the time after which its speed becomes half of initial is

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$ sec
• C. $1$ sec
• D. $\sqrt{\frac{3}{2}}$ sec

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}$ sec

Initial speed $u=10m.s$

Angle of projection $\theta ={60}^{\circ }$

Horizontal speed $u_x=u\cos {60}^{\circ }=10\times \frac{1}{2}=5m/s$

Speed will become half of initial at $t=\frac{T}{2}s$ when $v_y=0m/s$

$t=\frac{T}{2}=\frac{2u\sin \theta }{2g}=\frac{u\sin \theta }{g}=\frac{10\sin 60}{10}$

$=\frac{\sqrt{3}}{2}s$