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I am trying to solve this simple excercise:

Question

You throw a small coin upwards with $4 \frac{m}{s}$ . How much time does it need to reach the height of $0.5 m$ ? Why do we get two results?

Answer

(We get two results for time, because the coin passes the 0.5m mark downwards too.)

The equtation I used (for constant acceleration): $x = x_0 + v_0 t + \frac{1}{2} a t^2$

These values I know: $x = 0.5m$, $x_0 = 0$, $v_0 = 4 \frac{m}{s}$, $a = -9.81 \frac{m}{s^2}$ So, I get a quadratic equtation which I can solve for $t$. My two results are: $t_1 = 0.15s$ and $t_2 = 0.66s$, however the book has the results 0.804s and 0.013s. What am I doing wrong?

Thanks for your help

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closed as off-topic by tpg2114, Emilio Pisanty, Dimensio1n0, John Rennie, Chris White Nov 10 '13 at 10:14

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your answer is correct. wolframalpha.com/input/?i=solve+.5+%3D+4t+-(9.8/2)t^2 –  Approximist Mar 1 '11 at 22:00

1 Answer 1

up vote 2 down vote accepted

Nothing, your book did it wrong.

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2  
But if the question had asked for the times when a height of 0.05 m was reached, the book's answers would be correct. Sounds like a typo in the problem. –  Ted Bunn Mar 1 '11 at 22:04

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