$A$ body is projected from the earth at an angle of $30^{\circ}$ with the horizontal with some initial velocity. If its range is $20 \ m$, the maximum height reached by it is: (in meters)

$A$ particle is projected with a velocity $v$ such that its horizontal range is twice the greatest height attained. The horizontal range is

$A$ stone is projected with kinetic energy $E$,making an angle $\theta$ with the horizontal. When it reaches its highest point,its kinetic energy is

$A$ boy playing on the roof of a $10\, m$ high building throws a ball with a speed of $10\, m/s$ at an angle of $30^o$ with the horizontal. How far from the throwing point will the ball be at the height of $10\, m$ from the ground? $\left[ g = 10\, m/s^2, \sin 30^o = \frac{1}{2}, \cos 30^o = \frac{\sqrt{3}}{2} \right]$

$A$ cricket player hits a ball like a projectile,and the fielder catches the ball after $2 \ s$. The maximum height reached by the ball is $(g = 10 \ m/s^2)$. (in $m$)

If the maximum height and range of a projectile are $3 \,m$ and $4 \,m$ respectively,then the velocity of the projectile is (Take $g=10 \,ms^{-2}$)