$A$ projectile is thrown at an angle of $\pi/3$ with the horizontal,reaching a maximum height $Y$. If another projectile is thrown with the same initial velocity at an angle of $\pi/6$,what will be the maximum height attained?

The equation of a projectile is given by $y = Px - Qx^2$,where $P$ and $Q$ are constants. The ratio of the maximum height to the range of the projectile is

Two projectiles are thrown with the same initial velocity making an angle of $45^{\circ}$ and $30^{\circ}$ with the horizontal,respectively. The ratio of their respective ranges will be:

$A$ gun and a target are at the same horizontal level separated by a distance of $600 \,m$. The bullet is fired from the gun with a velocity of $500 \,ms^{-1}$. In order to hit the target, the gun should be aimed to a height $h$ above the target. The value of $h$ is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,m$)

If we can throw a ball up to a maximum height $H$,the maximum horizontal distance to which we can throw it is

$A$ ball is projected from the ground into the air. At a height of $5 \text{ m}$, its velocity is $\vec{v} = (5 \hat{i} + 5 \hat{j}) \text{ m s}^{-1}$. The maximum height reached by the ball is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$): (in $\text{ m}$)