Question

Algebra problem?

If a catapult is used to throw a lead ball, the path of the ball can be modeled by a quadratic equation, ax² + bx + c, where x is the horizontal distance that the ball travels and y is the height of the ball. If one of these catapult-launched lead balls travels 150 feet before hitting the ground and reaches a maximum height of 45 feet, what is the equation that represents its path?

Answer 1

`y=-3/375x^2+6/5x`

Explanation:

Let the maximum height be the vertex and let the maximum height be achieved at half of the total horizontal distance traveled:
`a(x-75)^2+45=y`

When `x=150`, `y=0`, solve for `a`
`a(150-75)^2+45=0`

`5625a=-45`

`a=-3/375`

`y= -3/375(x-75)^2+45`

`y= -3/375(x^2-150x+5625)+45`

`y= -3/375x^2+6/5x-45+45`

`y=-3/375x^2+6/5x`

graph{-3/375x^2+6/5x [-219.7, 219.8, -109.9, 109.8]}