$A$ 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 $\,m$)

$A$ cannon on a level plane is aimed at an angle $\theta$ above the horizontal and a shell is fired with a muzzle velocity $v_0$ towards a vertical cliff a distance $D$ away. Then the height from the bottom at which the shell strikes the side walls of the cliff is

$A$ projectile thrown with a speed $v$ at an angle $\theta$ has a range $R$ on the surface of the Earth. For the same $v$ and $\theta$,its range on the surface of the Moon will be:

$A$ particle aimed at a target,projected with an angle $15^{\circ}$ with the horizontal is short of the target by $10 \ m$. If projected with an angle of $45^{\circ}$ it is away from the target by $10 \ m$,then the angle of projection to hit the target is

$A$ shell is fired at an angle of $30^{\circ}$ to the horizontal with a velocity of $196 \,m/s$. What is the time of flight (in $\,s$)? (Take $g = 9.8 \,m/s^2$)

$A$ particle is projected from the ground at an angle of $\theta$ with the horizontal with an initial speed of $u$. The time after which the velocity vector of the projectile is perpendicular to the initial velocity is: