$A$ ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal from a distance $d_1$ from the foot of a pole. It just touches the top of the pole and then falls on the ground at a distance $d_2$ from the pole on the other side. The height of the pole is ...........

Define a projectile and derive the equation of its trajectory: $y = (\tan \theta_0)x - \frac{g}{2(v_0 \cos \theta_0)^2}x^2$.

$A$ 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. . . . .

The equations of motion of a projectile are given by $x = 36t$ metre and $2y = 96t - 9.8t^2$ metre. The angle of projection is:

The ratio of minimum kinetic energies of two projectiles of the same mass is $4:1$. The ratio of the maximum heights attained by them is also $4:1$. The ratio of their ranges would be: (in $:1$)

$A$ shell fired from the base of a mountain just clears it. If $\alpha$ is the angle of projection,then the angular elevation of the summit $\beta$ is