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 $^{\circ}$)?

$A$ stone is projected horizontally with a speed $10 \ m/s$ from an $80 \ m$ high building. The distance of the target on the ground from the foot of the building is $.... \ m$ $(g = 10 \ m/s^2)$.

$A$ 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 $\text{ m}$)

$A$ ground-to-ground projectile is at point $A$ at $t = \frac{T}{3}$,is at point $B$ at $t = \frac{5T}{6}$,and reaches the ground at $t = T$. The difference in heights between points $A$ and $B$ is

The equation of trajectory of a projectile is $y = 10x - (5/9)x^2$. If we assume $g = 10 \ m/s^2$, the range of the projectile (in meters) is:

In the given figure for a projectile,find the relation for $y$ in terms of $x_1, x_2$ and $\theta$.