$A$ projectile is projected with kinetic energy $K$. If it has the maximum possible horizontal range,then its kinetic energy at the highest point will be ......... $K$.

The horizontal range of a projectile is $4\sqrt{3}$ times its maximum height. Its angle of projection will be ......... $^\circ$.

$A$ projectile is given an initial velocity of $(3 \hat{i} + 4 \hat{j}) \text{ m s}^{-1}$, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. Assuming $g = 10 \text{ m s}^{-2}$, if the equation of its trajectory can be written as $\frac{1}{9} [\beta x + \gamma x^2]$, then the value of $\gamma$ is

$A$ body is projected at such an angle that the horizontal range is three times the greatest height. The angle of projection is

$A$ cricket fielder can throw a cricket ball with a speed $v_{0}$. If he throws the ball while running with speed $u$ at an angle $\theta$ to the horizontal,find:
$(a)$ The effective angle to the horizontal at which the ball is projected in the air as seen by a spectator.
$(b)$ The time of flight.
$(c)$ The horizontal range from the point of projection at which the ball will land.
$(d)$ The angle $\theta$ at which he should throw the ball to maximize the horizontal range found in $(c)$.
$(e)$ How does $\theta$ for maximum range change if $u > v_{0}$,$u = v_{0}$,and $u < v_{0}$?
$(f)$ How does $\theta$ in $(e)$ compare with that for $u = 0$ (i.e.,$45^{\circ}$)?

Assertion $(A)$: The range of a projectile is maximum when the angle of projection is $45^{\circ}$.
Reason $(R)$: The range of a projectile depends only on the angle of projection.