Simple kinematics excercise, throwing something upwards

Question

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

Answer 1

Nothing, your book did it wrong.

Answer 2

You get two results because the particle at this height does not stop but continues flying up. In a while it goes down and passes the same elevation on its way down.