$A$ bullet is fired from a cannon with a velocity of $500 \, m/s$. If the angle of projection is $15^\circ$ and $g = 10 \, m/s^2$,then the horizontal range is:

$A$ gun mounted on the ground fires bullets in all directions with same speed. The farthest distance the bullets could reach is $6.4 \text{ m}$. The speed of the bullets from the gun is . . . . . . m/s. (take $g = 10 \text{ m/s}^2$)

If a stone is to hit a point which is at a horizontal distance $d$ and at a vertical height $h$ above the point from where the stone is launched,what is the initial speed $u$ if the stone is launched at an angle $\theta$?

$A$ body of mass $m$ thrown up vertically with velocity $v_1$ reaches a maximum height $h_1$ in $t_1$ seconds. Another body of mass $2m$ is projected with a velocity $v_2$ at an angle $\theta$. The second body reaches a maximum height $h_2$ in time $t_2$ seconds. If $t_1 = 2t_2$,then the ratio $\left(\frac{h_1}{h_2}\right)$ is

$A$ particle is projected from the ground with a speed of $80 \,m/s$ at an angle of $30^{\circ}$ with the horizontal. The magnitude of the average velocity of the particle in the time interval $t=2 \,s$ to $t=6 \,s$ is ....... $m/s$. (Take $g=10 \,m/s^2$)

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: