Question

The height reached in time t by a partcle thrown upward with a speed u is given by h= ut - 1/2 gt^2
Find the time taken to reach maximum height.​

Answer 1

Without worrying about the derivative

dh/dt = 0,x this can be done by using

the quadratic.

The vertex is at (-b/2a, h(-b/2a))

h(t) = (-1/2)gt2 + ut

a = -g/2, b = u

-b/2a = -u/(-g) = u/g

h(u/g) = (-1/2)g(u/g)2 + u2/g

= -u2/2g + u2/g

= (-u2+2u2)/(2g)

= u2/2g

The maximum height is u2/2g

occurring at time t=u/g

hope helpful for you

Answer 2

Q. The height reached in time t by a partcle thrown upward with a speed u is given by $h = ut - \frac{1}{2} g {t}^{2}$

Find the time taken to reach maximum height.

$ans = > t = \frac{u}{g}$

Explanation:-

The height h is a function of time.Thus ,h will be maximum when dh/dt=0.

Given,

$h = ut - \frac{1}{2} g {t}^{2}$

or,

$\frac{dh}{dt} = \frac{d}{dt} (ut) - \frac{d}{dt} ( \frac{1}{2} g {t}^{2} )$

$= u \frac{dt}{dt} - \frac{1}{2} g \frac{d}{dt} ( {t}^{2} )$

$= u - \frac{1}{2} g(2t)$

$= u - gt$

For maximum h,

$\frac{dh}{dt} = 0$

or,

$u - gt = 0 \: \: or \: \: \: t = \frac{u}{g}$