Posted by **lola** on Friday, July 13, 2012 at 9:29pm.

a toy rocket is shot into the air at the top of a building so that its height above the ground t seconds after the rocket is launched is given by the formula h(t)= -3t cubed + 6t squared + 27t feet from liftoff until the rocket crashes into the ground. Find the maximum height height that the rocket achieves, the speed at which it crashes into the ground and the exact time it reached its maximum height.What is the instantaneous velocityat which the rocket nis thrown and two seconds after it is launched. What is the average velocityfor the first 2 seconds after it is launched. Alot I know but I cannot seem to catch onto the formulas!

- Math/calculus -
**Reiny**, Friday, July 13, 2012 at 11:45pm
h(t) = -3t^3 + 6t^2 + 27t

h' (t) = -9t^2 + 12t + 27

= 0 for a max/min of h(t)

divide by -3

3t^2 - 4t - 9 = 0

t = (4 ± √124)/6

= appr 2.523 or -1.18 , which we reject since time cannot be negative

h(2.523) = ..... you do the button-pushing.

initial velocity ---> t = 0

h' (0) = 27 ft/sec

when t=2

h'(2) = -9(4) + 12(2) + 27 = 15

average velocity in first 2 seconds:

h(0) = 0

h(2) = -3(8) + 6(4) + 27(2) = 54

avg vel = (54-0)/(2-0) = 27 ft/s

check my arithmetic

- Math/calculus -
**lola**, Saturday, July 14, 2012 at 11:20am
Thank you very much but how do I calculatethe speed at which it crashes to the ground? I got38.181 for the max height do I start from there?

Thanks again, going through the calculations you gave makes it easier to understand how I use the formulas. I am never sure when to use first derivatives, 2nd etc-when to factor etc you have been very helpful.

lola

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