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

A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3\, ms^{-2}$ for $ 0.5\, minutes$. If the maximum height reached by it is $80\, m$, then the angle of projection is (Take $g = 10\, ms^{-2}$)

A ball is thrown from a point with a speed ${v_o}$ at an angle of projection $\theta $. From the same point and at the same instant a person starts running with a constant speed ${v_o}/2$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection

From the ground level, a ball is to be shot with a certain speed. Graph shows the range $(R)$ of the particle versus the angle of projection from horizontal ( $\theta $ ). Values of $\theta _1$ and $\theta _2$ are

Choose the correct alternative $(s)$

The maximum vertical height to which a man can throw a ball is $136\,m$. The maximum horizontal distance upto which he can throw the same ball is $.....\,m$