The vertical displacement ($y$ in metre) of a projectile in terms of its horizontal displacement ($x$ in metre) is given by $y = (\sqrt{3}x - 0.2x^2)$. The time of flight of the projectile is (Acceleration due to gravity $g = 10 \ ms^{-2}$)

If we can throw a ball up to a maximum height $H$,the maximum horizontal distance to which we can throw it is

The equation of trajectory of a projectile is $y = 10x - (5/9)x^2$. If we assume $g = 10 \ m/s^2$,the range of the projectile (in meters) is:

The horizontal and vertical displacements $x$ and $y$ of a projectile at a given time $t$ are given by $x = 6t \text{ m}$ and $y = 8t - 5t^2 \text{ m}$. The range of the projectile in metres is:

$A$ stone is projected from the ground with velocity $25\,m/s$. Two seconds later,it just clears a wall $5\,m$ high. The angle of projection of the stone is ........ $^o$ $(g = 10\,m/s^2)$.

$A$ 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.,$45^{\circ}$)?