Chapter 9, Problem 28SGR

To determine

To find: Theequation of the parabola that models the flight of the football.

Answer to Problem 28SGR

The equation of parabola is $y\text{ =}\frac{1}{25}{\left(x-50\right)}^2+100$ .

Explanation of Solution

Given information:

A football is thrown from ground level. It reaches a maximum height of $50$ feet and hits the ground $200$ feet from where it was thrown.

Calculation:

A football is thrown from ground level. It reaches a maximum height of $50$ feet and hits the ground $200$ feet from where it was thrown. The path of the football is a parabola such that the football was kicked at the origin.

The parabola is of the form $y\text{ =a}{\left(x-h\right)}^2+k$ . The point of maximum height (A) will be the vertexof the parabola.

  

The horizontal distance from point O to point A will be half of the distance of OB. So,

  $OC=100ft$

Hence the co-ordinates of vertex A will be

  $\left(h,k\right)=\left(50,100\right)$

Substitute $h=\text{ 5}0$ and $k=100$ in the standard form $y\text{ =a}{\left(x-h\right)}^2+k$ the equation of the parabolic path of the baseball will be,

  $y\text{ =a}{\left(x-50\right)}^2+100\mathrm{..........}(1)$

Parabola passes through point B. Co-ordinates of point Bare $\left(0,200\right)$ Hence

  $\begin{array}{c}200=a{\left(0-50\right)}^2+100\\ 200=2500a+100\\ 2500a=100\end{array}$

  $a=\frac{1}{25}$

Substitute $a=\frac{1}{25}$ in equation (1) $y\text{ =a}{\left(x-50\right)}^2+100$ , the equation of the parabola that models the flight of the football will be

  $y\text{ =}\frac{1}{25}{\left(x-50\right)}^2+100$

Therefore, the equation of parabola is $y\text{ =}\frac{1}{25}{\left(x-50\right)}^2+100$ .