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

$A$ body is projected with a speed $u \ m/s$ at an angle $\beta$ with the horizontal. The kinetic energy at the highest point is $3/4$ of the initial kinetic energy. The value of $\beta$ is: (in $^{\circ}$)

$A$ bullet is fired at time $t=0$ with velocity $20 \ m/s$ and at an initial angle of $30^{\circ}$ with the horizontal. The angle between the displacement vector and the horizontal after time $0.1 \ s$ is (Assume $g=10 \ m/s^2$)

$A$ particle aimed at a target,projected with an angle $15^{\circ}$ with the horizontal,is short of the target by $10 \ m$. If projected with an angle of $45^{\circ}$,it is away from the target by $15 \ m$. Then the angle of projection to hit the target is:

For a projectile,the ratio of the maximum height reached to the square of the time of flight is:

$A$ small ball is thrown at an angle $45^{\circ}$ to the horizontal with an initial velocity of $2 \sqrt{2} \ m/s$. The magnitude of the mean velocity averaged over the first $2 \ s$ is (take acceleration due to gravity $g = 10 \ m/s^2$). (in $m/s$)