Question

Two second after projection, a projectile is travelling in a direction inclined at ${30}^{\circ }$ to the horizontal. After 1 more
Second, it is travelling horizontally. Then $(g=10m/s^2)$

• A. the velocity of projection is $20\sqrt{3}m/s$
• B. the angle of projection is ${30}^{\circ }$ with horizontal
• C. Both (A) and (B) are correct
• D. Both (A) and (B) are wrong

Answer 1

The correct option is A the velocity of projection is $20\sqrt{3}m/s$

$\frac{T}{2}=2+1=3sorT=6s$
$\therefore \frac{2u_y}{g}=6$
$\therefore u_y=30m/s$
Further, $tan{30}^{\circ }=\frac{v_y}{v_x}=\frac{u_y-gt}{u_x}=\frac{30-20}{u_x}$
or $u_x=10\sqrt{3}m/s$
or $u=\sqrt{u_x^2+u_y^2}=20\sqrt{3}m/s$
$tan\theta =\frac{u_y}{u_x}=\frac{30}{10\sqrt{3}}=\sqrt{3}$
or $\theta ={60}^{\circ }$