$A$ stone is thrown at an angle $\theta$ to the horizontal reaches a maximum height $H$. Then the time of flight of the stone will be

If the range of a body projected with a velocity of $60 \,m \,s^{-1}$ is $180 \sqrt{3} \,m$, then the angle of projection of the body is (Acceleration due to gravity $= 10 \,m \,s^{-2}$)

$A$ player kicks a ball at a speed of $20 \ ms^{-1}$ so that its horizontal range is maximum. Another player $24 \ m$ away in the direction of the kick,starts running in the same direction at the same instant the ball is kicked. If he has to catch the ball just before it reaches the ground,he should run with a velocity equal to: (Take $g = 10 \ ms^{-2}$)

$A$ projectile is thrown at a speed which is twice its speed at its maximum height. If $R$ and $H$ are its range and maximum height respectively, then the ratio $\frac{R}{H}$ is

The equation of trajectory of a projectile is $y = 10x - (5/9)x^2$. If we assume $g = 10 \ m/s^2$, the range of the projectile (in meters) is:

The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is $x$. The angle of projection with the horizontal is