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 $g = 10 \ ms^{-2}$)

$A$ ground-to-ground projectile is at point $A$ at $t = \frac{T}{3}$,is at point $B$ at $t = \frac{5T}{6}$,and reaches the ground at $t = T$. The difference in heights between points $A$ and $B$ is

The equations of motion of a projectile are given by $x = 36t \, m$ and $2y = 96t - 9.8t^2 \, m$. The angle of projection is

$A$ ball is thrown from a point on the ground at some angle of projection. At the same time,a bird starts from a point directly above this point of projection at a height $h$ and moves horizontally with speed $u$. Given that in its flight,the ball just touches the bird at one point. Find the distance on the ground where the ball strikes.

If the velocity at the maximum height of a projectile projected at an angle of $45^{\circ}$ is $20 \,m \,s^{-1}$, then the maximum height reached by the projectile is (Acceleration due to gravity $=10 \,m \,s^{-2}$ ) (in $\,m$)

$A$ swimmer dived off a cliff with a running horizontal leap. What must his minimum speed be just as he leaves the top of the cliff so that he will miss the edge at the bottom,given that the ledge at the bottom is $2 \ m$ wide and $10 \ m$ below the top of the cliff?