$A$ cricketer hits a ball with a velocity $25\,m/s$ at $60^\circ$ above the horizontal. How far above the ground does it pass over a fielder $50\,m$ from the bat (in $,m$)? (Assume the ball is struck very close to the ground and $g = 9.8\,m/s^2$)

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5t^2) \ m$ and $x = 6t \ m$,where $t$ is in seconds. The angle with the horizontal at which the projectile was projected is:

$A$ projectile is thrown with a velocity of $10 \sqrt{2} \ m/s$ at an angle of $45^{\circ}$ with the horizontal. The time interval between the moments when the speed is $\sqrt{125} \ m/s$ is $(g = 10 \ m/s^2)$. (in $s$)

If the velocity of a projectile at its maximum height is $\frac{1}{\sqrt{2}}$ times its initial velocity,what is its range?

If a body projected with a velocity of $19.6 \ m/s$ reaches a maximum height of $9.8 \ m$,then the range of the projectile is (Neglect air resistance). (in $m$)

The time of flight of an object projected with speed $20 \, m/s$ at an angle $30^{\circ}$ with the horizontal is .... $s$.