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Using Equations for Constant Acceleration- Vertical Motion of a Bullet

  1. Jan 16, 2013 #1
    1. The problem statement, all variables and given/known data

    A bullet is shot vertically into the air at a speed of 512 m/s.

    a) to what maximum height does the bullet go?

    b) how much time passes before the bullet stops rising?

    c) what is the velocity of the bullet after 60s?

    2. Relevant equations

    a) V2²=V1²+2ad

    b) a= (V2-V1)/Δt

    c) V2=V1+aΔt

    3. The attempt at a solution

    a) Given: V2: 0 m/s; V1: 512 m/s; a: 9.8 m/s²
    Unknown: distance
    Equation: d=(V2²+V1²)/2a
    Substitute: d=(0 m/s + 512 m/s)/2(9.8 m/s²)
    Solve 26.1 m
    ∴ the bullet will reach a maximum height of 26.1 m.

    b) Given: V2: 0 m/s; V1: 512 m/s; a: 9.8 m/s²
    Unknown: time
    Equation: Δt=(V2-V1)/a
    Substitute: Δt=(0 m/s-512 m/s)/9.8 m/s²
    Solve: 52.2s
    ∴ the bullet will stop rising after 52.2s

    c) Given: t: 60s; a: -9.8m/s²; V1: 512 m/s
    Unknown: V2
    Equation: V2=V1+aΔt
    Substitute: V2=512 m/s + (-9.8 m/s²)(60s)
    Solve: 76m/s
    ∴ the bullet's velocity after 60s is 76m/s

    I would like for someone to please look over my work and tell me if it's correct or not. Thank you.
  2. jcsd
  3. Jan 16, 2013 #2


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    heck, i could throw a baseball that high at a mere fraction of that initial speed...you forgot to square it..:wink:
    looks good!
    that is its speed...what is its velocity?
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