The height reached in time t by a particle thrown upwared with a speed u is given by
`h=ut-1/2 gt^2`
where g=9.8 m/s^2 is a constant.Find the timetakenin raching the maximum height.

Explanation

The height h is a function of tiem.Thus, h will be maximum when `(dh)/(dt)=0` We have,
`h=ut-1/2 gt^2`
or, `(dh)/(dt)=d/(dt) (ut)-d/dt(1/2 gt^2)`
`=u(dt)/(dt)-/2g d/(dt) (t^2)`
`= u-1/2 g(2g)=u-gt.` For maximum h,
`(dh)/(dt)=0`
or, `u-gt=0 or, t=u/g`