Q. A particle is projected up with an initial velocity of $80\ ft/sec$. The ball will be at a height of $96\ ft$ from the ground after
A) $2.0$ and $3.0\ sec$
B) Only at $3.0\ sec$
C) Only at $2.0\ sec$
D) After $1$ and $2\ sec$
Explanation
(a) $h = ut - \frac{1}{2}gt^{2} \Rightarrow 96 = 80t - \frac{32}{2}t^{2}$
$\Rightarrow t^{2} - 5t + 6 = 0 \Rightarrow t = 2sec$ or $3\ sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.