Two spheres are projected at angles $30^{\circ}$ and $45^{\circ}$ with the horizontal. The maximum height reached by both is same. The ratio of their initial velocities is, $(\sin 45^{\circ} = \frac{1}{\sqrt{2}}, \sin 30^{\circ} = 0.5)$

$A$ 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 $g = 9.8\,m/s^2$)

$A$ ball is projected from the ground into the air. At a height of $5 \text{ m}$, its velocity is $\vec{v} = (5 \hat{i} + 5 \hat{j}) \text{ m s}^{-1}$. The maximum height reached by the ball is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$): (in $\text{ m}$)

$A$ projectile can have the same range for two angles of projection. If $h_1$ and $h_2$ are the maximum heights when the range in the two cases is $R$,then the relation between $R$,$h_1$,and $h_2$ is:

In a projectile motion,the velocity at maximum height is:

From the top of a tower, $80 \ m$ high from the ground, a stone is thrown in the horizontal direction with a velocity of $8 \ ms^{-1}$. The stone reaches the ground after a time $t$ and falls at a distance of $d$ from the foot of the tower. Assuming $g=10 \ ms^{-2}$, the time $t$ and distance $d$ are given respectively by