Question 312206: a model rocket is shot into the air and its path is apporoximated by h=-5t^2 + 30t, where h is the height of the rocket above the ground in metres and t is the elapsed time in seconds. When will the rocket hit the ground and what is the maximum height of the rocket.
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! a model rocket is shot into the air and its path is apporoximated by h=-5t^2 + 30t, where h is the height of the rocket above the ground in metres and t is the elapsed time in seconds. When will the rocket hit the ground and what is the maximum height of the rocket.
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To find out when it will hit the ground, set h to zero and solve for t:
h=-5t^2 + 30t
0=-5t^2 + 30t
0=-5t(t - 5)
t = {0, 5}
We can throw out the 0 because this is when it was first launched. Therefore:
rocket will hit the ground in 5 seconds
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To find max height, find the "axis of symmetry":
x = -b/2a
x = -30/2(-5)
x = -30/(-10)
x = 3 sec
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Plug above into equation to find height:
h=-5t^2 + 30t
h=-5*3^2 + 30(3)
h=-5*9 + 30(3)
h=-45 + 90
h= 45 feet
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