you throw ball over 10ft fence. ball reaches highest point directly above fence & lands 10 ft from fence???

Question

using fence as axis of symmetry, what is quadratic function that models ball's height????????????????????

Answer 1

Obviously at the starting point (-10), the height is zero. At the fence (0), the height is ten. And at a point past the fence (10), the height is again 0.

This is an upside down parabola going through the points (0,10) and (10,0).

If you start with a parabola f(x) = x², this goes the wrong direction, so make it negative.

Upside down would be:
f(x) = -x²

But if you graph this, it is all below the x-axis. We want it to pass through (0,10) and (10,0). How can we do that? Let's add a constant.

f(x) = -x² + k

Now try some values:
f(0) = -(0)² + k

Ah! k should be 10 because at the top of the arc it is just at the top of the fence.

f(x) = -x² + 10

Okay, but now we have a problem because f(10) doesn't equal 0. Let's multiply the first term times n.

f(10) = -n(10)² + 10

n = 1/10

So the final equation is:
f(x) = -(1/10)x² + 10

Answer 2

Let the quadratic be y = ax^2 + bx + c where the origin is at the base of the fence.
The curve goes through (0, 10) (the highest point) and through (10, 0) where it lands.

The highest point is where dy/dx = 2ax + b = 0. If it's at x = 0, 2a(0) + b = 0, so b = 0.
In the original curve, 10 = a(0)^2 + 0(0) + c, so c = 10.

To find a, substitute x = 10, y = 0:
0 = 100a + 0 + 10. So a = 1/10.

Answer 3

first find the horizonal speed of the ball when it reaches its maximum height.

To do that, you need to find the time it takes the ball the hit the ground

h(t) = -16t^2 + hi
0 = -16t^2 + 10
-10 = -16t^2
t^2 = 10/16
t = 0.79056s

the time it takes the ball to hit the ground vertically is also the time it takes the ball to land horizonally

x = vt
10 = .79056v
v = 12.64926 ft/s

x = 12.64926t
t = x/12.64926

h(t) = -16t^2 + hi

plug x/12.64926 for t

h(t) = -16(x/12.64926)^2 + 10 <== answer

EDIT: i rounded the answer so it's alittle off.
10 = sqrt(5/8)v
v = 10/sqrt(5/8)

so t = 10x/sqrt(5/8)

t = 10x/sqrt(5/8)

h(t) = -16 (10x/ sqrt(5/8))^2 + 10
h(t) = -16x^2 / (160) + 10
h(t) = -x^2/10 + 10 <== if i used the exact value