a model rocket is launched vertically into the air such that its height at any time, t, is given by the function h(t)=-16t^2+80t+10. What is the maximum height attained by the model rocket?
A 140
B 110
C 85
D 10
Answers
Answer:
a) set h(t) = 0 and solve for t
-16t^2 + 128t = 0 implies 16t^2 = 128t. Dividing both sides by 16t yields the equation t = 8
It will take 8 seconds after launch for the rocket to return to the ground.
b) set h(t) = 112 and solve for t. 16t^2 -128t + 112 = 0. Divide both sides by 16 to obtain t^2 -8t + 7 = 0.
This can be factored into (t-7)(t-1)=0. Therefore, the rocket is 112 feet above the ground at t=1 and also t=7.
There are two times when the rocket is 112 feet above the ground, once after 1 second (on the way up), and once after 7 seconds (on the way down.)
c) To find the maximum height, set the first derivative equal to zero. The first derivative of the equation 16t^2 - 128t + 112 = 0 is
32t -128 = 0 which implies t = 4
The rocket will reach its maximum height 4 seconds after launch.
d) simply use the answer from part c0 above in the original equation. h(t) = -16t^2 + 128t
Now, h(t) = -16(4)^2 + 128(4) = -16*16 + 512 = -256 + 512 = 256
The maximum height is 256 feet.
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Step-by-step explanation: