If the height of a projectile at a time of $2 \ s$ from the beginning of motion is $60 \ m$,then the time of flight of the projectile is (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $s$)

Neglecting air resistance,the time of flight of a projectile is determined by:

$A$ body of mass $10 \ kg$ is projected at an angle of $45^{\circ}$ with the horizontal. The trajectory of the body is observed to pass through a point $(20, 10)$. If $T$ is the time of flight,then its momentum vector,at time $t = \frac{T}{\sqrt{2}}$,is. [Take $g = 10 \ m/s^2$]

If the initial kinetic energy of a projectile is $100 \, J$ and the kinetic energy at the maximum height is $30 \, J$,what is the angle of projection?

$A$ gun and a target are at the same horizontal level separated by a distance of $600 \,m$. The bullet is fired from the gun with a velocity of $500 \,ms^{-1}$. In order to hit the target, the gun should be aimed to a height $h$ above the target. The value of $h$ is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,m$)

$A$ bullet fired from a gun falls at a distance half of its maximum range. The angle of projection of the bullet is (in $^{\circ}$)