Question

A ball is projected with a velocity $20\sqrt{3}m/s$ at angle ${60}^o$ to the horizontal. The time interval after which the velocity vector will make an angle ${30}^o$ to the horizontal is (in seconds)

• A. $3$
  
• B. $2$
  
• C. $5$
  
• D. $9$
  

Answer 1

The correct option is B $2$

Given : $u=20\sqrt{3}m/s$ $\theta ={60}^o$
Horizontal component of velocity initially $u_x=u\cos 60=20\sqrt{3}\times \frac{1}{2}=10\sqrt{3}m/s$
Since, there is no acceleration in horizontal direction. So horizontal component of velocity remains the same all the time.
Final horizontal component of velocity $v_x=u_x=10\sqrt{3}m/s$
Final velocity makes an angle ${30}^o$ with horizontal.
So, vertical component of final velocity $v_y=u_x\tan 30=10\sqrt{3}\times \frac{1}{\sqrt{3}}=10$
Initial vertical comonent of velocity $u_y=u\sin 60=20\sqrt{3}\times \frac{\sqrt{3}}{2}=30m/s$
Using $v_y=u_y-gt$
$\therefore$ $10=30-10t$
$⟹t=2s$