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?

Question

1015 views around the world

You can reuse this answer
Creative Commons License

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!