The horizontal range $(R)$ of a projectile is $n$ times its maximum height $(H)$. Find the angle of projection $(\theta_0)$.

Assertion $(A)$: The velocity of a projectile at a point on its trajectory is equal to the slope at that point.
Reason $(R)$: The velocity vector at a point is always along the tangent to the trajectory at that point.

$A$ body of mass $1 \, kg$ is projected from the ground at an angle of $30^{\circ}$ with the horizontal on level ground at a speed of $50 \, m/s$. The magnitude of the change in momentum of the body during its entire flight is ....... $kg \cdot m/s$ $(g = 10 \, m/s^2)$.

$A$ ball is projected from the ground with a speed of $20\,m/s$ at an angle of $45^{\circ}$ with the horizontal. There is a wall of $25\,m$ height at a distance of $10\,m$ from the projection point. The ball will hit the wall at a height of $.........\,m$.

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

$A$ 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?