question_answer

A ball is dropped from the top of \[80\text{ }m\]high tower. If after 2 s of fall the gravity \[(g=10\,m/{{s}^{2}})\] disappears, then time taken to reach the ground since the gravity disappeared is

A) \[2\,s\]              

B) \[3\,s\]

C) \[4\,s\]              

D) \[5\,s\]

Correct Answer: B

Solution :

Velocity acquired by the ball after 2 s of drop is \[v=gt=10\times 2=20m/s\] 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 \[t'3\,\,s\].