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Maximum height of a spring toy

  1. Apr 26, 2016 #1
    1. The problem statement, all variables and given/known data
    A spring toy is launched from the ground at 2.3 m/s at an angle of 78° to the ground. What is the maximum height reached by the spring toy?

    2. Relevant equations

    Δ
    dv = v1Δt + 1/2aΔt2


    3. The attempt at a solution

    Calculate time:

    Δt = 2.3 m/s / 9.8m/s2

    = 0.23 s

    Now calculate vertical motion:
    Δdv = v1Δt + 1/2aΔt2 *for vertical motion, let (up) be positive

    Δdv = (2.3 m/s) x (0.23s) + 1/2 (-9.8 m/s2) x (0.23s)2

    Δdv = 0.27 m [up]

    The maximum height reached by the spring toy is 0.27 m [up].

    Is this correct? I have a feeling I did it wrong because I did not use the angle they provided and only used the velocity.
     
  2. jcsd
  3. Apr 26, 2016 #2

    SteamKing

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    No, it's not correct. Why didn't you calculate the vertical component of the initial velocity?
     
  4. Apr 26, 2016 #3
    Okay, so I did it again with the vertical component of initial velocity.

    Calculate vertical component of initial velocity:

    V1v = v1sinθ

    V1v = (2.3 m/s) sin 78°

    V1v = 2.25 m/s [up]

    Calculate time:

    Δt = 2.3 m/s / 9.8m/s2

    = 0.23 s

    Calculate vertical motion:
    Δdv = v1Δt + 1/2aΔt2 *for vertical motion, let (up) be positive

    Δdv = (2.25 m/s) x (0.23s) + 1/2 (-9.8 m/s2) x (0.23s)2

    Δdv = 0.26 m [up]

    The maximum height reached by the spring toy is 0.26 m [up].

    Is this better?
     
  5. Apr 26, 2016 #4

    SteamKing

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    Much better.
     
  6. Apr 26, 2016 #5
    Thanks for the help :)
     
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