$A$ ball projected from the ground at an angle of $45^o$ just clears a wall in front. If the point of projection is $4 \, m$ from the foot of the wall and the ball strikes the ground at a distance of $6 \, m$ on the other side of the wall,the height of the wall is ........ $m$.

$A$ bullet is dropped from a certain height,and at the same instant,another bullet is fired horizontally from the same height. Which bullet will hit the ground first?

$A$ projectile is thrown with an initial velocity $(a \hat{i} + b \hat{j}) \ m/s$,where $\hat{i}$ and $\hat{j}$ are unit vectors along horizontal and vertical directions respectively. If the range of the projectile is twice the maximum height reached by it,then

$A$ 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 $\text{ m}$)

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

$A$ stone is projected vertically upwards with speed '$v$'. Another stone of same mass is projected at an angle of $60^{\circ}$ with the vertical with the same speed '$v$'. The ratio of their potential energies at the highest points of their journey is $\left[\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]$