# Question #288b9

May 24, 2017

4 seconds.

#### Explanation:

Damn, this kid has got some seriously strong legs.

graph{-16x^2+64x [-72.2, 75.9, -4, 70]}

Comment aside, I think we should rewrite the function to best suit the current circumstance:
$f \left(t\right) = - 16 {t}^{2} + 64 t$ where $t$ is the time in seconds.
Now, we have a function of time. This graph above shows the relationship between the ball's height $f \left(t\right)$ and the time that has passed $t$. In this case, when $f \left(t\right) = 0$ the ball is on the ground, as the height $f \left(t\right)$ is zero.

Alright, so all we need to do is to find the difference in time. The difference in time, let's call it $\Delta t$ is the distance between the two x-intercepts (really t-intercepts should be more accurate, see graph).
$\Delta t = {t}_{2} - {t}_{1}$

And the x-intercepts can be found using the quadratic formula:
$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

With $a = - 16 , b = 64 , c = 0$
$t = 0 \mathmr{and} 4$
$\Delta t = 4$

$\therefore$ The time it takes for the ball to hit the ground is 4 seconds.