An object is thrwon vertically upwards with a velocity of 60 m/s. After what time it strike the ground ? Use `g=10m//s^(2)`.

Explanation

Let the time of ascent of the particle be t
using `v=u-g t`
`t=(-u)/(-g)=(60)/(10)=6s`
As the time of ascent is equal to the time of descent, therefore
`t=6+6=12s`