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## Homework Statement

An arrow is shot vertically upward with an initial speed of 25 m/s. Two seconds later, another arrow is shot upward with the same speed. On its way up, the second arrow meets the first arrow on its way down. Assume that both arrows experience a constant downward acceleration of 9.8 m/s

^{2}.

How long does it take (from the time the first arrow is shot) for the arrows to meet?

## Homework Equations

d = vf * t - 0.5(a)(t)

^{2}

t

_{A1}= t

_{A2}+ 2

## The Attempt at a Solution

d

_{A1}= 25t - 4.9t

^{2}

d

_{A2}= 25t - 4.9t

^{2}

Since d

_{A1}= d

_{A2}

25t

_{A1}- 4.9t

_{A1}

^{2}= 25t

_{A2}- 4.9t

_{A2}

^{2}

Since t

_{A1}= t

_{A2}+ 2

25(t

_{A2}+ 2) - 4.9(t

_{A2}+ 2)

^{2}= 25t

_{A2}- 4.9t

_{A2}

^{2}

19.6t = 50-19.6

t = 1.55 s

The correct answer is 3.55 s