Question

A particle is projected up with an initial velocity of $80ft/\sec$. The ball will be at a height of $96ft$ 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

Answer 1

Answer: (A)

2.0 and 3.0 sec

Answer 2

$h = ut - \frac{1}{2}gt^{2} \Rightarrow 96 = 80t - \frac{32}{2}t^{2}$

$\Rightarrow t^{2} - 5t + 6 = 0 \Rightarrow t = 2$sec or 3 sec