SOLUTION: s=vt+16t2 object released from 684 feet the initial velocity is 18 ft/s and air resistance is neglected how many seconds will the object hit the ground?

Question 926001: s=vt+16t2
object released from 684 feet the initial velocity is 18 ft/s and air resistance is neglected how many seconds will the object hit the ground?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
s=vt+16t2
object released from 684 feet the initial velocity is 18 ft/s and air resistance is neglected how many seconds will the object hit the ground?
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Use s(t) = -16t^2 + vt + so
If the in+tial velocity is downward (you didn't say) it's -18. If it's upward, it's +18.
I'll assume downward.
so is the starting height.
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s(t) = -16t^2 - 18t + 684 = 0
Solve for t

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=44100 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -7.125, 6. Here's your graph:

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Ignore the negative solution.
t = 6 seconds
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