Question

A ball is projected with a velocity $20\sqrt{2}m{s}^{-1}$ 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 (Take $g=10m{s}^{-2})$

• A. $5s$
• B. $2s$
• C. $1s$
• D. $3s$

Answer 1

The correct option is B $2s$

$U_x=20\sqrt{3}\times \frac{1}{2}=10\sqrt{3}m/s$

$U_y=20\sqrt{3}\times \frac{\sqrt{3}}{2}=30m/s$

$\Rightarrow V_x=U_x=V\cos {30}^o$

$\Rightarrow V_y=U_y-gt=V\sin {30}^o$

$\frac{V_y}{V_x}=\tan {30}^o=\frac{30-10t}{10\sqrt{3}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{30-10t}{10\sqrt{3}}$

$10t=20$

$t=2$