Question

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?

Answer 1

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 `(t+3)` to reach this point while the second a time `t`.
We use the relationship of kinematics:
`y=v_it+1/2at^2`
Where `v_i` is the initial velocity and `a` is the acceleration of gravity.
So we have:

(I rounded a little bit....)