# An archer shoots an arrow with an initial velocity of 21 m/s straight up from his bow. He quickly reloads and shoots another arrow the same way 3 s later. At what time and height do the arrows meet?

Dec 31, 2014

The first arrow and the second meet at a height y.

#### Explanation:

$t = 3.64 s$
$y = 11.4 m$

The first arrow and the second meet at a height $y$.

The first one takes a time $\left(t + 3\right)$ to reach this point while the second a time $t$.
We use the relationship of kinematics:
$y = {v}_{i} t + \frac{1}{2} a {t}^{2}$
Where ${v}_{i}$ is the initial velocity and $a$ is the acceleration of gravity.
So we have:

(I rounded a little bit....)