Question

A soccer ball is kicked off from the ground in an arc defined by the function, h(x)=-8x^2+64x. At what point does the ball hit the ground?
(0,4) , (0,8) , (4,0) , (8,0)

Answer 1

Answer:

(8,0)

Step-by-step explanation:

The equation that models the path traced by the ball is

$h(x)=-8x^2+64x$

To find the point at which the ball hit the ground, we must equate the function to zero.

$-8x^2+64x=0$

Factor;

$-8x(x-8)=0$

$-8x=0,(x-8)=0$

This implies that;

x=0,x=8,

At x=0, the ball was not yet kicked.

So we take x=8, to be the time the ball hit the ground.

We substitute x=8 into the function to get;

$h(8)=-8(8)^2+64(8)=0$

Hence the point at which the ball hit the ground is (8,0)