Better Students Ask More Questions.
The height of lava ejected can be modeled by h=-5t(t-11) where h is height in meters...
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.
Posted by jeew-m on June 23, 2013 at 2:30 AM (Answer #2)
Differentiating with respect to t, we get
h'`= -5*2t+55 = -10t+55`
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.
Posted by llltkl on June 23, 2013 at 2:16 AM (Answer #1)
Join to answer this question
Join a community of thousands of dedicated teachers and students.