# Simple kinematics excercise, throwing something upwards [closed]

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?

(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?

-

## closed as off-topic by tpg2114, Emilio Pisanty, centralcharge, John Rennie, Chris WhiteNov 10 '13 at 10:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – tpg2114, Emilio Pisanty, centralcharge, John Rennie, Chris White
If this question can be reworded to fit the rules in the help center, please edit the question.

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