$A$ projectile covers double the range compared to its maximum height attained. The angle of projection is

If the velocity of a projectile at its maximum height is $\frac{1}{\sqrt{2}}$ times its initial velocity,what is its range?

$A$ ball is thrown at $30^{\circ}$ with the horizontal from the top of a roof $20 \,m$ high with a speed of $13 \,ms^{-1}$. At what distance from the throwing point will the ball, once again, be at a height of $20 \,m$ from the ground (in $\,m$)? $(g = 10 \,ms^{-2})$

$A$ body is projected at an angle $\theta$ with the horizontal. Another body is projected with the same speed at an angle $\theta$ with the vertical. The ratio of their maximum heights is:

$A$ ball is projected with a velocity $5 \text{ m/s}$, such that its horizontal range is twice the greatest height attained. The value of the range is (in $\text{ m}$)

The maximum vertical height to which a man can throw a ball is $136\,m$. The maximum horizontal distance up to which he can throw the same ball is $.....\,m$.