$A$ ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected with the same speed at an angle of $30^{\circ}$ with the horizontal. At the highest point,the ratio of their potential energies is

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range $R$. If the time of flight of these bodies are $5 \text{ s}$ and $10 \text{ s}$, respectively, then the value of $R$ is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$)

$A$ player can throw a ball to a maximum horizontal distance of $80 \,m$. If he throws the ball vertically with the same velocity, then the maximum height reached by the ball is (in $\,m$)

Two paper screens $A$ and $B$ are separated by $150 \ m$. $A$ bullet pierces $A$ and then $B$. The hole in $B$ is $15 \ cm$ below the hole in $A$. If the bullet is traveling horizontally at the time of hitting $A$,then the velocity of the bullet at $A$ is: $(g=10 \ m \ s^{-2})$

$A$ particle is projected at an angle of $60^{\circ}$ with the horizontal from the ground with a velocity $10 \sqrt{3} \ m/s$. The angle between the velocity vector after $2 \ s$ and the initial velocity vector is $(g = 10 \ m/s^2)$. (in $^{\circ}$)

$A$ projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 \, ms^{-2}$ for $0.5 \, minutes$. If the maximum height reached by it is $80 \, m$,then the angle of projection is (Take $g = 10 \, ms^{-2}$)