$A$ particle is projected with velocity $2 \sqrt{gh}$,so that it just flies over two walls of equal height $h$ and $2h$ distance apart from each other. Find the time for which the particle flies between the walls.

$A$ ball is projected with kinetic energy $E$, at an angle of $60^{\circ}$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will be:

$A$ particle is thrown with a velocity of $u \, m/s$ at an angle of $30^{\circ}$ with the horizontal. It passes through points $A$ and $B$ at the same vertical height at times $t_1 = 1 \, s$ and $t_2 = 3 \, s$ respectively,as shown in the figure. The value of $u$ is ....... $m/s$ (take $g = 10 \, m/s^2$).

$A$ ball is thrown from the location $(x_0, y_0) = (0, 0)$ of a horizontal playground with an initial speed $v_0$ at an angle $\theta_0$ from the $+x$-direction. The ball is to be hit by a stone,which is thrown at the same time from the location $(x_1, y_1) = (L, 0)$. The stone is thrown at an angle $(180^{\circ} - \theta_1)$ from the $+x$-direction with a suitable initial speed $v$. For a fixed $v_0$,when $(\theta_0, \theta_1) = (45^{\circ}, 45^{\circ})$,the stone hits the ball after time $T_1$,and when $(\theta_0, \theta_1) = (60^{\circ}, 30^{\circ})$,it hits the ball after time $T_2$. In such a case,$(T_1 / T_2)^2$ is. . . . .

$A$ projectile $A$ is thrown at an angle $30^{\circ}$ to the horizontal from point $P$ with velocity $v_1$. At the same time,another projectile $B$ is thrown with velocity $v_2$ upwards from point $Q$,which is vertically below the highest point that $A$ would reach. For $B$ to collide with $A$,the ratio $\frac{v_2}{v_1}$ should be:

$A$ particle is projected from the ground with a kinetic energy $E$ at an angle of $60^{\circ}$ with the horizontal. Its kinetic energy at the highest point of its motion will be