$A$ stone is projected horizontally with a speed $10 \ m/s$ from an $80 \ m$ high building. The distance of the target on the ground from the foot of the building is $.... \ m$ $(g = 10 \ m/s^2)$.

$A$ body is projected horizontally from a tower with a velocity of $18 \, ms^{-1}$. It strikes the ground at an angle of $45^o$. The vertical component of the velocity when it strikes the ground is ........ $ms^{-1}$.

$A$ projectile is thrown at an angle $\theta$ such that it is just able to cross a vertical wall of height $H$ at its highest point,as shown in the figure. The horizontal distance from the point of projection to the wall is $\sqrt{3} H$. The angle $\theta$ at which the projectile is thrown is given by:

$A$ particle is projected with a velocity $v$ such that its horizontal range is twice the greatest height attained. The horizontal range is

$A$ 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. . . . .

If a ball can be thrown to a maximum horizontal distance of $80 \, m$,what is the maximum height to which it can be thrown (in $, m$)?