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. The initial height of the second ball is $(g = {10ms}^{- 2})$
A) 6.25 m
B) 2.5 m
C) 3.75 m
D) 5 m
Explanation
(b) Maximum height of projectile
$H = \frac{u^{2}\sin^{2}\theta}{2g}$
$\therefore\ \ H = \frac{(10)^{2} \times \sin^{2}(30{^\circ})}{2 \times 10}$
$= \frac{5}{4} = 1.25\ m$
Time to reach maximum height
$t = \frac{usin\theta}{g}$
$\therefore\ \ t = \frac{10 \times sin\ 30{^\circ}}{10} = 0.5s$
So, distance of vertical fall in 0.5s
$h = \frac{1}{2}gt^{2}$
$or\ \ h = \frac{1}{2} \times 10 \times (0.5)^{2} = 1.25\ m$
$\therefore Height\ of\ second\ ball = 1.25 + 1.25 = 2.5\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.