The velocity at the maximum height of a projectile is $\frac{\sqrt{3}}{2}$ times its initial velocity of projection $(u)$. Its range on the horizontal plane is .............

For a given velocity,a projectile has the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then:

$A$ cricket ball thrown across a field is at heights $h_{1}$ and $h_{2}$ from the point of projection at times $t_{1}$ and $t_{2}$ respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is

$A$ ball $P$ is projected at an angle of $60^{\circ}$ with the vertical with a certain initial speed $u$. Another ball $Q$ of the same mass $m$ as that of ball $P$ is projected vertically upwards with the same initial speed $u$. At the highest point, the ratio of the potential energy of ball $P$ to that of ball $Q$ is $(\sin 30^{\circ} = 0.5)$.

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 $m$)

$A$ projectile is projected with a velocity of $25 \, m/s$ at an angle $\theta$ with the horizontal. After $t$ seconds,its inclination with the horizontal becomes zero. If $R$ represents the horizontal range of the projectile,the value of $\theta$ will be: [use $g = 10 \, m/s^2$]