# Freefall Arrow Problem: Finding the initial speed and maximum height

1. Sep 14, 2012

### Parkkk41

1. An arrow is shot vertically upward with an initial speed of 25 m/s. When it's exactly halfway to the top of its flight, a second arrow is launched vertically upward from the same spot. The second arrow reaches the first arrow just as the first arrow reaches its highest point. What is the launch speed of the second arrow?

2. v2 = v1 + a(Δt), Δd = v2(Δt)-(1/2)a(Δt)^2, Δd = v1(Δt)+(1/2)a(Δt)^2

3. I determined the known values in order to find the time and distance of the first arrow:
v1= 25 m/s
a= -9.8 m/s^2 (u)
v2 = 0 m/s
Δd= ?
Δt= ?

I then found Δt by using the formula v2= v1+a(Δt) and got 2.55 s. Next, I found the acceleration by using Δd=v2Δt-(1/2)a(Δt)^2 which was 31.86 m... I figured I could find the time of the second arrow to the same height that I found but I'm not sure how to go about that. Is it half the time or is there a way I should be calculating it? Very confused! This is just a homework question but it's really bugging me that I don't know how to do it. Any help is appreciated! Thank you :)

Last edited by a moderator: Sep 24, 2012
2. Sep 14, 2012

### PeterO

These objects moving vertically are traveling faster near the ground, so the time taken to reach half height is less than the time taken for the second half of the flight.
This arrow will take less than one second to reach half height, despite taking over 2.5 second to reach maximum height.
To get the time time of the second arrow, you have to calculate the time taken for the first arrow to each half height.
You again use your formula Δd=v2Δt-(1/2)a(Δt)^2, but this time you know Δd and want to calculate Δt.
It is a quadratic equation so you will get two answers, but remember that after passing through this height, the arrow goes to full height, then comes back down again.

Last edited by a moderator: Sep 24, 2012