$A$ large number of bullets are fired in all directions with the same speed $u$. The maximum area on the ground on which the bullets will spread is

$A$ ball is projected from a building of height $10 \ m$ with a velocity of $10 \ m/s$ at an angle of $30^\circ$ with the horizontal. What is the horizontal distance covered by the ball when it reaches the same height of $10 \ m$ again (in $m$)? $(g = 10 \ m/s^2, \sin 30^\circ = 1/2, \cos 30^\circ = \sqrt{3}/2)$

$A$ particle is thrown with a speed $u$ at an angle $\theta$ with the horizontal. When the particle makes an angle $\phi$ with the horizontal,its speed changes to $v$,where

$A$ 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:

$A$ ball at point '$O$' is at a horizontal distance of $7 \ m$ from a wall. On the wall,a target is set at point '$C$'. If the ball is thrown from '$O$' at an angle $37^{\circ}$ with the horizontal,aiming at the target '$C$',but it hits the wall at point '$D$',which is at a vertical distance '$y_0$' below '$C$'. If the initial velocity of the ball is $15 \ m/s$,find $y_0$ (given $\cos 37^{\circ} = \frac{4}{5}$). (in $m$)

The angle of projection for a projectile to have the same horizontal range and maximum height is