Why are there two answers to (b)?

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SUMMARY

The discussion centers on the physics problem of a stone thrown vertically upward at a speed of 24.0 m/s, specifically addressing the time required to reach a height of 13.0 m. The solution reveals that there are two distinct times when the stone reaches this height: once while ascending and once while descending. This phenomenon occurs because the stone passes through the 13 m mark twice during its trajectory, except at the apex where it momentarily stops before falling back down.

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Homework Statement



A stone is thrown vertically upward with a speed of 24.0 m/s.

a) How fast is it moving when it reaches a height of 13.0 m/s?
b) How much time is required to reach this height?
c) Why are there two answers to (b)?

Homework Equations


[/B]
x = v0t +- 1/2 at2

The Attempt at a Solution



I had no problem with the first two, but I've thought I need an explanation for the third one. Here is the graph and solutions. Does the second root appear because of the stone's falling right after reaching its maximum height?

http://www4c.wolframalpha.com/Calculate/MSP/MSP42811iec8be4feb3fd0e00001afd1ed1gd449fc6?MSPStoreType=image/gif&s=16&w=86.&h=18. http://www4c.wolframalpha.com/Calculate/MSP/MSP42841iec8be4feb3fd0e00003d457bh4d7i14gi8?MSPStoreType=image/gif&s=16&w=78.&h=18.
http://www4c.wolframalpha.com/Calculate/MSP/MSP42881iec8be4feb3fd0e0000509big5995b3732e?MSPStoreType=image/gif&s=16&w=392.&h=192.&cdf=Coordinates&cdf=Tooltips
 
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That's right. What goes up must come down!

It passes the 13m level twice. In fact, it passes through every level twice, except for the point at the very apex of the trajectory.
 
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NascentOxygen said:
What goes up must come down!
Unless it exceeds escape velocity :D
 

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