A ball of mass `m_(1)` and another ball of mass `m_(2)` are dropped from equal height. If the time taken by the balls are `t_(1)` and `t_(2)`, respectively, then

Options

A. `t_(1)=(t_(2))/(2)`
B. `t_(1)=t_(2)`
C. `t_(1)=4t_(2)`
D. `t_(1)=(t_(2))/(4)`

Answer

b

Explanation

For mass `m_(1)`, time taken to fall from height h is
`t_(1)=(sqrt(2h)/(g))`
For mass `m_(2)`
`t_(2)=sqrt((2h)/(g))`
`t_(1)=t_(2)`