Question

A balloon going upward with a velocity of $ 12 \,m/s $ is at a height of $ 65\, m $ from the earth's surface at any instant. Exactly at this instant a ball drops from it. How much time will the ball take in reaching the surface of earth ? $ (g= 10\, m/s^2) $

• A. $ 5\,s $
• B. $ 6\,s $
• C. $ 10\,s $
• D. None of these

Answer 1

Answer: (A)

From equation of motion, we have
$h=u t+\frac{1}{2} g t^{2}$
where $u$ is initial velocity, $g$ the acceleration due to gravity and $t$ the time taken.
Taking upward direction as negative and downward direction as positive, we have
$h=65\, m$,
$u=-12 \,m / s , g=10 \,m / s ^{2}$

$\therefore \, 65=-12 t+\frac{1}{2} \times 10 \times t^{2}$
$\therefore \, 5 t^{2}-12 t-65=0$
$\Rightarrow \,(t-5)(5 t+13)=0$
$\therefore \, t=5\, s$