Question 266702: Hi everyone, I'm having a bit of a problem with, well, a problem. We've been dealing with quadratic equations and I'm pretty sure that's how to solve the problem, however I can't figure out how to set up the problem.
Here's the posed problem:
In March of 2000, Doug Danger made a sucessful motorcycle jump over an L-1011 jumbo jet. The horizontal distance of his jump was 160 feet, and his height, in feet, during the jump was approximated by h = -16t^2 + 25.3t + 20, t>/=0.
He left the takeoff ramp at a height of 20 feet and he landed on the landing ramp at a height of about 17 feet. How long, to the nearest 10th of a second, was he in the air.
Now I did attempt to solve for T by setting h to zero and h to 17.
Setting h to zero gave me -25.3 +/- sqrt (25.3^2-4(-16)(20)/2(-16) = -25.3 +/- sqrt 1920.09/-32
Setting h to 17 = -25.3 +/- sqrt 448.09/-32
I'm obviously missing something, but what?
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! "He left the takeoff ramp at a height of 20 feet and he landed on the landing ramp at a height of about 17 feet"
he takes off at time zero ___ this is why h=20 when t=0
solving for t when h=17 gives the flight time
___ there is also a negative t solution for h=17 ___ this roughly corresponds to time BEFORE takeoff, when he is going up the ramp
17 = -16t^2 + 25.3t + 20 ___ 0 = -16t^2 + 25.3t + 3 ___ b^2 - 4ac = 832.09
using quadratic formula ___ t = -.11 and t = 1.69 ___ 1.7 to the nearest tenth
"I'm obviously missing something, but what?"
___ it looks like the sign is reversed in the b^2 - 4ac of your calculation for h=17
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