# A soccer ball kicked at the goal travels in a path given by the parametric equations: x=50t; y=-16t^2+32t, Suppose the ball enters the goal at a height of 5ft. How far away from the goal was the kicker?

Apr 9, 2017

The kicker was $50 \left(1 \pm \frac{\sqrt{11}}{4}\right)$ ft away from the goal.

#### Explanation:

Let us find time $t$ when the ball is at a height of 5 ft.

By setting $y = 5$,

$- 16 {t}^{2} + 32 t = 5 R i g h t a r r o w - 16 {t}^{2} + 32 t - 5 = 0$

$t = 1 \pm \frac{\sqrt{11}}{4}$

If the ball enters the goal on the way up, then the kicker was

$x = 50 \left(1 - \frac{\sqrt{11}}{4}\right)$ ft away from the goal.

If the ball enters the goal on the way down, then the kicker was

$x = 50 \left(1 + \frac{\sqrt{11}}{4}\right)$ ft away from the goal.

I hope that this was clear.