SOLUTION: A model rocket is launched from the roof of a building with height h0. For height h, in meters, and time t, in seconds, after the rocket is launched, the height of the rocket above
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Question 1082791: A model rocket is launched from the roof of a building with height h0. For height h, in meters, and time t, in seconds, after the rocket is launched, the height of the rocket above the ground is given by
h=f(t)=−5t^2+50t+40.
Report any noninteger answers accurate to 2 decimal places.
Note: Due to the dynamic nature of this question, the graph may not be exactly to scale.
(a) What is the height, in meters, of the building?
(b) When will the rocket hit the ground?
(c) When will the rocket be 25 meters in the air?
(d) Your response Correct response
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
a) The rocket is at the top of the building the instant it is launched, that is when 
. So what is the value of )
?
b) The height of the ground is zero, so what is the positive value of 
that makes \ =\ 0)
?
c) Since the rocket starts out higher than 25 meters, the rocket will be at 25 meters at the positive value of that makes \ =\ 25)
Hint: for b) and c) you will get both positive and negative values of . But since the rocket is launched at , any negative value of would represent time BEFORE the rocket was launched -- information that is of no use.
John
My calculator said it, I believe it, that settles it
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