- #1

- 2

- 0

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 :)

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: