At what angle of elevation should a projectile be projected with a velocity of $20 \, m/s$ to reach a maximum height of $10 \, m$ (in $^{\circ}$)?

Three identical balls are projected with the same speed at angles $30^{\circ}$,$45^{\circ}$,and $60^{\circ}$. Their ranges are $R_1$,$R_2$,and $R_3$ respectively. Then:

From the top of a tower $19.6 \,m$ high, a ball is thrown horizontally. If the line joining the point of projection to the point where it hits the ground makes an angle of $45^{\circ}$ with the horizontal, then the initial velocity of the ball is (in $\,m/s$)

$A$ projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 \, ms^{-2}$ for $0.5 \, minutes$. If the maximum height reached by it is $80 \, m$,then the angle of projection is (Take $g = 10 \, ms^{-2}$)

$A$ ball $P$ is dropped vertically and another ball $Q$ is thrown horizontally with the same initial vertical velocity (which is $0$) from the same height $h$ at the same time. If air resistance is neglected,then:

$A$ ball is thrown at an angle $\theta$ and another ball is thrown at an angle $(90^\circ - \theta)$ with the horizontal from the same point with the same speed $40\,m/s$. The second ball reaches $50\,m$ higher than the first ball. Find their individual heights.