Find the horizontal distance the ball covers.Assume the height is zero when the ball is kicked and when the ball is caught.

Question

h(X)=-1/2(x-20)^2 +50

Answer 1

h(x) = -(1/2)(x-20)^2 +50

 

For h=0,

 

-(1/2)(x-20)^2 +50 = 0

x^2 - 40x + 400 - 100 = 0

x^2 - 40x + 300 = 0

 

x = (1/2){40±√{(40)^2 - 4(300)] = (1/2)(40±20) = 20±10 = 30 or 10

 

The ball is kicked up from the ground at x=10, then reaches the ground again at x=30, so its horizontal travel is 30-10=20