Paper airplane follows the path #y=-2x^2+20x+1# where y represents the height of the paper airplane in feet and x represents the seconds it has traveled. what is the time before the airplane will reach 15 feet?

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Answer 1 · 2016-05-10T23:48:46

15 is the value of y, so we will solve as we would a regular quadratic equation.

15 = -2x^2 + 20x + 1#

#0 = -2x^2 + 20x - 14#

x = #(-b +- sqrt(b^2 - 4ac))/(2a)#

#x = (-20 +- sqrt(20^2 - 4 xx -2 xx -14))/(2 xx -2)#

#x = (-20 +- sqrt(288))/-4#

#x = 0.757 or 9.243#

Therefore, the paper airplane will be at 15 feet 0.757 seconds and 9.243 seconds after its launch.

Hopefully this helps!