# A projectile is thrown upward so that its distance above the ground after t secondes is h=-12t^2+432t. After how many seconds does it reach its maximum height?

*print*Print*list*Cite

### 3 Answers

If you have quadratic function (parabola) `f(x)=ax^2+bx+c` where `a<0` (as in your case) its maximum value will be `-(b^2-4ac)/(4a)` (that is the height of vertex of parabola), and that maximum value will be reached at point `-b/(2a)` (those are seconds in your case).

So in your case `a=-12,` `b=432` and `c=0.`

**Hence projectile will reach its maximum height after** `-432/(2cdot(-12))=18` **seconds.**

That maximum value will be `h=-12cdot18^2+432cdot18=3888`

Red is maximum height, green is time in seconds till maximum height and blue is trajectory of projectile.

We have given

h=-12t^2+432t

`h=-12t^2+432t`

at maximum height , its velocity will be zero.So differentiate h with respect to t ,so we have

`h'(t)=-24t+432`

`h'(t)=0 if `

-24t+432=0

t=18 sec.

Also

`h''(t)=-24 <0`

Thus

t=18 Sec , give maximum height.

We have given

h=-12t^2+432t

`h(t)=-12t^2+423t`

at maximum height , its velocity will be zero.So differentiate h with respect to t

`h'(t)=-24t+423`

`h'(t)=0 if`

`-24t+423=0`

`t=423/24`

`t=17.63 sec.`

`h''(t)=-24 <0`

`Thus`

t=17.63 Sec give maximum height.