The initial speed of a projectile fired from the ground is $u$. At the highest point during its motion,the speed of the projectile is $\frac{\sqrt{3}}{2} u$. The time of flight of the projectile is:

$A$ ground-to-ground projectile is at point $A$ at $t = \frac{T}{3}$,is at point $B$ at $t = \frac{5T}{6}$,and reaches the ground at $t = T$. The difference in heights between points $A$ and $B$ is

What do you mean by projectile motion and projectile particle? Find the value of the position of a projectile particle at any instant of time.

$A$ body is projected at an angle $\theta$ so that its range is maximum. If $T$ is the time of flight,then the value of maximum range is (acceleration due to gravity $= g$)

$A$ projectile can have the same range for two angles of projection. If $h_1$ and $h_2$ are the maximum heights when the range in the two cases is $R$,then the relation between $R$,$h_1$,and $h_2$ is:

$A$ projectile thrown from the ground has initial speed $u$ and its direction makes an angle $\theta$ with the horizontal. If at maximum height from the ground,the speed of the projectile is half its initial speed of projection,then the maximum height reached by the projectile is:
$[g = \text{acceleration due to gravity}, \sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2]$