The height of lava ejected can be modeled by h=-5t(t-11) where h is height in meters and t is time in seconds since it was ejected. Find max height

### 2 Answers | Add Yours

The max height can be found by completing the square.

`h = -5t(t-11)`

`h = -5(t^2-11t)`

`h = -5(t^2-11t+121/4-121/4)`

`h = -5((t-11/2)^2-121/4)`

`h = -5(t-11/2)^2+605/4`

The max height will be given when `-5(t-11/2)^2 ` is minimum or in other words when it is 0.

Max h = 605/4m = 151.25m

*So the max height is 151.25m.*

**Sources:**

Given that,

h=-5t(t-11)

=-5t^2+55t

Differentiating with respect to t, we get

h'`= -5*2t+55 = -10t+55`

h"`= -10`

At the maximum point of any function, h’=0; h”=-ve

Putting h’=0, -10t+55 = 0

`rArr` t = 55/10 = 5.5 s

h” is negative there too.

Therefore, maximum height is reached at time t = 5.5 s.

Height attained at t=5.5 s is,

`h_(max) = -5 (5.5)^2+55*5.5`

`=151.25 ` meters.

Therefore, maximum height of lava from the point of its ejection is 151.25 meters.

**Sources:**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes