If we can throw a ball up to a maximum height $H$,the maximum horizontal distance to which we can throw it 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.,$45^{\circ}$)?

$A$ body is projected from the earth at an angle of $30^{\circ}$ with the horizontal with some initial velocity. If its range is $20 \ m$,the maximum height reached by it is: (in meters)

If $x$ and $y$ coordinates of a projectile as a function of time are given as $24t$ and $43.6t - 4.9t^2$,respectively,then the angle (in degrees) made by the projectile with horizontal when $t = 2 \text{ s}$ is . . . . . . .

$A$ stone projected from the ground with a velocity $50 \ m/s$ at an angle of $30^{\circ}$ with the horizontal crosses a wall after a time of $3 \ s$. Then the horizontal distance beyond the wall that the stone strikes the ground is (acceleration due to gravity $g = 10 \ m/s^2$) (in $m$)

$A$ particle is projected from level ground. Its kinetic energy $K$ changes due to gravity such that $\frac{K_{\max}}{K_{\min}} = 9$. The ratio of the range $R$ to the maximum height $H$ attained during its flight is: