The trajectory of a projectile is given by $y = 12x - \frac{3}{4}x^2$. If the horizontal component of velocity is $3 \ m/s$, find the range of the projectile. $(g = 10 \ m/s^2)$ (in $m$)

$A$ 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 $m$)? $(g = 10 \ m/s^2, \sin 30^\circ = 1/2, \cos 30^\circ = \sqrt{3}/2)$

For a projectile motion, the range $R$ is '$n$' times the maximum height $H$. So the angle of projection is

$A$ ball projected at an angle of $45^{\circ}$ with the horizontal crosses two points at equal heights separated by a distance at times $2 \ s$ and $8 \ s$ respectively. The horizontal distance between the two points is (Acceleration due to gravity $= 10 \ m/s^2$) (in $m$)

$A$ 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

The velocity at the maximum height of a projectile is half of its initial velocity $u$. Its range on the horizontal plane is