Q. A ball is projected up at an angle $\theta$ with horizontal from the top of a tower with speed $v$. It hits the ground at point $A$ after time $t_{A}$ with speed $v_{A}$ . Now, this ball is projected at same angle and speed from the base of the tower (located at point $P$) and it hits ground at point $B$ after time $t_{B}$ with speed $v_{B}$. Then
J & K CETJ & K CET 2015Motion in a Plane
Solution:
Consider, the two projectiles projected from $O$ and $P$ as shown in the figure.
Range of both the particles $R=\frac{v^{2} \sin 2 \theta}{g}$
$\Rightarrow O A'=P B=R$ Clearly, velocity at the points $A'$ and $B$ will be same. Velocity of the ball following the trajectory OC'A'. Will increase after A' due to accelerated motion under gravity. Therefore, we can say that $v_{A}>v_{B}$
