Question #6c732

1 Answer
C. McCracken
Apr 21, 2017

Sorry, what's the question here? 6 seconds, if you're looking for the time it takes for the rock to hit the ground.

Explanation:

It's not super clear what you're asking (no offence).

If you're asking how long it takes the rock to hit the ground (just my guess) I can answer that.

We know the height is given as
h(t)=-16x^2+16x+480h(t)=−16x2+16x+480

though I feel like the "x"s were supposed to be "t"s, since h was a function of t and you didn't mention t in the equation. To rewrite,

h(t)=-16t^2+16t+480h(t)=−16t2+16t+480

We also know that when the rock hits the ground,
h(t)=0h(t)=0.

So we need to solve the quadratic for when h(t)=0h(t)=0

0=-16t^2+16t+4800=−16t2+16t+480
We can divide by 16
0/16=-t^2+t+30016=−t2+t+30
0=-t^2+t+300=−t2+t+30

Now we can factor this or use the quadratic formula . Just for fun, let's use the quadratic formula. I'm assuming you know how to solve quadratics, but if not perhaps check out the links above.

t=\frac{-1 +- \sqrt(1^2-4(-1)(30))}{2(-1)}t=−1±√12−4(−1)(30)2(−1)
=\frac{-1 +- \sqrt(121)}{-2}=−1±√121−2
=\frac{1 +- \sqrt(121)}{2}=1±√1212
=\frac{1 +- 11}{2}=1±112
=\frac{12}{2} or -10/2=122or−102
=6 or -5=6or−5
But the negative 5 doesn't really make sense in this case (the equation doesn't apply for t less than 0), so let's take the value of 6.

So, after 6 seconds, the rock will hit the ground.

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