Question

The height reached in time t by a particle thrown upward with a speed u is given by
`h=ut-(1)/(2) "gt"^(2)`
where g is acceleration due to gravity. Find the time taken in reaching the maximum height.

Answer 1

Here h=f(t).
Thus h will be maximum when `(dh)/(dt)=0`
We have `h=ut- (1)/(2) "gt"^(2)`
or `(dh)/(dt)=u-(1)/(2)g(2t)=u-"gt"`
for maximum h, `(dh)/(dt)=0`
`rArr " " u-"gt"=0`
`t=(u)/(g)`