If the velocity at the maximum height of a projectile projected at an angle of $45^{\circ}$ is $20 \,m \,s^{-1}$, then the maximum height reached by the projectile is (Acceleration due to gravity $=10 \,m \,s^{-2}$ ) (in $\,m$)

An aeroplane flying at a constant velocity releases a bomb. As the bomb drops down from the aeroplane,

The horizontal range of a projectile projected at an angle of $45^{\circ}$ with the horizontal is $50 \ m$. The height of the projectile when its horizontal displacement is $20 \ m$ is (in $m$)

The horizontal and vertical displacements of a projectile at time $t$ are $x=36 t$ and $y=48 t-4.9 t^2$, respectively. Initial velocity of the projectile in $m/s$ is

Two balls with same mass and initial velocity are projected at different angles in such a way that the maximum height reached by the first ball is $8$ times higher than that of the second ball. If $T_1$ and $T_2$ are the total flight times of the first and second ball,respectively,then the ratio of $T_1$ to $T_2$ is:

At which points of its trajectory will a projectile have minimum and maximum speed?