Question #288b9

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Question #288b9

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Answer 1 · 2017-05-24T01:24:33

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:
$f (t) = - 16 t^{2} + 64 t$ $f(t)=-16t^2+64t$ where $t$ $t$ is the time in seconds.
Now, we have a function of time. This graph above shows the relationship between the ball's height $f (t)$ $f(t)$ and the time that has passed $t$ $t$ . In this case, when $f (t) = 0$ $f(t)=0$ the ball is on the ground, as the height $f (t)$ $f(t)$ is zero.

Alright, so all we need to do is to find the difference in time. The difference in time, let's call it $Deltat$ is the distance between the two x-intercepts (really t-intercepts should be more accurate, see graph).
$Deltat=t_2-t_1$

And the x-intercepts can be found using the quadratic formula:
$t=(-b+-sqrt(b^2-4ac))/(2a)$

With $a=-16,b=64,c=0$
$t=0 or 4$
$Deltat=4$

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

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Question #288b9

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