Asked by Joel

A stunt motorcyclist makes a jump from one ramp 20 feet off the ground. The jump between ramps can be modeled by y= -1/640x^2 + 1/4x +20 where x is the horizontal distance (in feet) and y is the height above ground (in feet).
I think a is 20 feet, but I don't know how to do b, c, or d. Please show me how to do the problems.
A. What is the motorcyclists height when he lands on the ramp?
B. What is the distance d between the ramps?
C. What is the horizontal distance h the motorcyclist has traveled when it reaches its maximum height?
D. What is the motorcyclists maximum height k above the ground.
Answered by Reiny
A) If the two ramps have the same height, then your are correct at 20 ft. The height as he leaves the ramp is 20 , so hopefully he lands on the 2nd ramp at a height of 20 ft

B) set y = 20

- 1/640x^2 + (1/4)x + 20 = 20
multiply by -640
x^2 - 160x = 0
x(x - 160) = 0
x = 0 or x = 160

so the distance from take-off to landing is from 0 to 160.
So the distance between them is 160 ft

C) the max height would be at the vertex , that is, when the x = 80
so he reaches the highest point at the 80 ft mark

d) to get the max height, plug in x = 80 into the original equation
I will let you do that, let me know what you get.
Answered by Joel
D is 30 feet above the ground.
Answered by Reiny
correct, that is what I had
Answered by Sydney
30
Answered by Matthew
Honestly god bless these people who ask these exact math questions almost a decade before you do. Really saving me right now.
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