$A$ ball having kinetic energy $KE$ is projected at an angle of $60^{\circ}$ from the horizontal. What will be the kinetic energy of the ball at the highest point of its flight?

At the top of the trajectory of a projectile,the acceleration is

$A$ bomb is dropped by an airplane flying horizontally with a velocity of $200 \text{ km/hr}$ at a height of $980 \text{ m}$. At the time of dropping the bomb,the horizontal distance of the airplane from the target on the ground to hit it directly is (given $g = 9.8 \text{ m/s}^2$):

$A$ 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 $m/s$)

$A$ particle is projected from the ground at an angle of $\theta$ with the horizontal with an initial speed of $u$. The time after which the velocity vector of the projectile is perpendicular to the initial velocity is:

$A$ 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 $g = 10 \, ms^{-2}$)