Two boys are standing at the ends $A$ and $B$ of a ground where $AB=a$. The boy at $B$ starts running in a direction perpendicular to $AB$ with velocity $v_1$. The boy at $A$ starts running simultaneously with velocity $v$ and catches the other boy at time $t$, when $t$ is:

Two particles are projected with the same initial velocity, one making an angle $\theta$ with the horizontal and the other making an angle $\theta$ with the vertical. If their common range is $R$, then product of time of flight is directly proportional to:

A particle is projected with speed $u$ so as to strike at the right angle a plane through the point of projection inclined at $30°$ to the horizon. Show that the range on this inclined plane is:

A particle $P$ is sliding down a frictionless hemispherical bowl, Fig.  It passes the point $A$ at $t=0$. At this instant of time, the horizontal component of its velocity is $v$. A bead $Q$ of the same mass as $P$ is ejected from $A$ at $t=0$ along the horizontal string $AB$, with the speed $v$. Friction between the bead and the string may be neglected. Let $t_{\mathrm{P}}$ and $t_Q$ be the respective times taken by $P$ and $Q$ to reach the point $B$. Then:

The Fig. shows the displacement-time graph for a particle. Find the time duration $\left(0-ts\right)$for which the average velocity of the particle is zero.