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 $3.0sec$
• B. Only at $3.0sec$
• C. Only at $2.0sec$
• D. After $1$ and $2sec$

Answer 1

The correct option is A $2.0$ and $3.0sec$

Given:

$u=80ft/sec$

$h=96ft$

Now, $h=ut-\frac{1}{2}g{t}^2$

$96=80t-\frac{1}{2}\times 32\times t^2$

$6=5t-t^2$

$t^2-5t+6=0$

$(t-2)(t-3)=0$

$t=2secor3sec$