$A$ particle is projected at an angle of $45^{\circ}$ with the horizontal with an initial kinetic energy $K$. What will be its kinetic energy at the highest point?

$A$ helicopter is flying horizontally with a speed $v$ at an altitude $h$ and has to drop a food packet for a man on the ground. What is the distance of the helicopter from the man when the food packet is dropped?

For angles of projection of a projectile at angles $(45^\circ + \theta)$ and $(45^\circ - \theta)$,the horizontal range described by the projectile are in the ratio of:

$A$ particle of mass $m$ is projected with a velocity $u$ making an angle $45^o$ with the horizontal. The magnitude of the torque due to the weight of the projectile,when the particle is at its maximum height,about the point of projection is:

$A$ body is projected at an angle $\theta$ so that its range is maximum. If $T$ is the time of flight, then the value of maximum range is (acceleration due to gravity $= g$)

For a projectile motion, the range $R$ is '$n$' times the maximum height $H$. So the angle of projection is