# A soccer ball kicked at the goal travels in a path given by the parametric equations: x=50t; y=-16t^2+32t, At what two times will the ball be at a height of 6ft.?

Jul 15, 2017

time $= 0.2094 , 1.7906 \setminus \setminus \setminus \left(\text{unit}\right) \setminus \setminus \left(4 \mathrm{dp}\right)$

#### Explanation:

We have:

$x = 50 t$
$y = - 16 {t}^{2} + 32 t$

On the assumption that $t$ represents time, and $x$ and $y$ are the horizontal and vertical displacements respectively in feet then we note that:

Horizontal displacement is linear
Vertical displacement is parabolic

So for the vertical displacement to be $6$ feet we require:

$- 16 {t}^{2} + 32 t = 6$
$\therefore 16 {t}^{2} - 32 t + 6 = 0$

And using the quadratic formula we form two solutions:

$t = 1 \pm \frac{1}{2} \sqrt{\frac{5}{2}}$
$\setminus \setminus = 0.2094 , 1.7906 \setminus \setminus \left(4 \mathrm{dp}\right)$