$A$ projectile with speed $50 \ m \ s^{-1}$ is thrown at an angle of $60^{\circ}$ with the horizontal. The maximum height that can be reached is,(acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m$)

$A$ ball at point '$O$' is at a horizontal distance of $7 \ m$ from a wall. On the wall,a target is set at point '$C$'. If the ball is thrown from '$O$' at an angle $37^{\circ}$ with the horizontal,aiming at the target '$C$',but it hits the wall at point '$D$',which is at a vertical distance '$y_0$' below '$C$'. If the initial velocity of the ball is $15 \ m/s$,find $y_0$ (given $\cos 37^{\circ} = \frac{4}{5}$). (in $m$)

An object is projected at an angle of $45^\circ$ with the horizontal. The ratio of the horizontal range to the maximum height reached is:

$A$ particle is projected at $60^{\circ}$ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is

$A$ body is projected at an angle of $45^{\circ}$ with the horizontal with a velocity of $60 \sqrt{2} \ m/s$. Then the angle made by its velocity with the horizontal after $6 \ s$ is (in $^{\circ}$)

$A$ bomb of mass $m$ is projected from the ground with speed $v$ at an angle $\theta$ with the horizontal. At the maximum height from the ground,it explodes into two fragments of equal mass. If one fragment comes to rest immediately after the explosion,then the horizontal range of the centre of mass is ..............