Q. A ball is dropped from the top of $80\, m$ high tower. If after $2 \sec$ of fall the gravity $\left(g=10\, m / s ^{2}\right)$ disappears, then time taken to reach the ground since the gravity disappeared is
J & K CETJ & K CET 2015Motion in a Straight Line
Solution:
Velocity acquired by the ball after $2\, s$ of drop is
$v=g t=10 \times 2=20\, m / s$ Height fallen by the ball is
$h=\frac{1}{2} g t^{2}=\frac{1}{2} \times 10 \times 4=20\, m$
After gravity disappears, speed is constant.
$\Rightarrow H-h=(v) \times t'$
$\Rightarrow 80-20=(20) \times t'$
$\Rightarrow 60=20\, t'$ or $t'3 \,s.$
