Q. A ball is projected from the ground at a speed of $10ms^{- 1}$ making an angle of $30{^\circ}$ with the horizontal. Another ball is simultaneously released from a point on the vertical line along the maximum height of the projectile. Both the balls collide at the maximum height of first ball. The initial height of the second ball is $(g = 10ms^{- 2})$
A) $6.25m$
B) $2.5m$
C) $3.75m$
D) $5m$
Explanation
(b) Maximum height of projectile, $h_{0} = \frac{u^{2}\sin^{2}\theta}{2g}$
$\therefore h_{0} = \frac{(10)^{2} \times \sin^{2}{30{^\circ}}}{2 \times 10} = \frac{5}{4} = 1.25\ m$
Time for attaining maximum height , $t = \frac{u\sin\theta}{g}$
$\therefore t = \frac{10 \times \sin{30{^\circ}}}{10} = 0.5\ sec$
$\therefore$ Distance of vertical fall in 0.5 sec, $S = \frac{1}{2}gt^{2}$
$\Rightarrow S = \frac{1}{2} \times 10 \times (0.5)^{2} = 1.25m$
$\therefore$ Height of second ball$\ = 1.25 + 1.25 = 2.50m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.