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Projectile Motion- Object Launched in the Air

  1. Apr 19, 2017 #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
    ## \vec v_1v = \vec v_1 sin\Theta##

    ## \Delta \vec d = \vec v_1v \Delta t + \frac 1 2 \vec a \Delta t^2##

    3. The attempt at a solution

    ## \vec v_1v = 2.3 m/s [up] (sin78 degrees)
    \\ = 2.25 m/s [up]##

    ## \Delta \vec d = \vec v_v1 \Delta t + \frac 1 2 \vec a \Delta t^2
    \\ 0 = (2.25 m/s [up]) \Delta t + \frac 1 2 (-9.8 m/s^2) \Delta t^2
    \\ 0 = \Delta t (2.25 m/s - 4.9 m/s^2) \Delta t
    \\ \Delta t = 0.46 seconds##

    That's the total time it was airborne. Maximum height is half that so 0.46/2 = 0.23 s.

    Now to find the height.

    ## \Delta \vec v = \vec v_1 \Delta t + \frac 1 2 \vec a \Delta t^2
    \\ = (2.25 m/s) (0.23 s) + \frac 1 2 (-9.8 m/s^2) (0.23)^2
    \\ = 0.52 - 0.26
    \\= 0.26 m [up]##

    The max height is 0.26 meters [up].

    I've followed my textbook's lead of rounding to two significant places through the work.

    Am I correct at all?
     
  2. jcsd
  3. Apr 19, 2017 #2

    scottdave

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    Everything looks good to me. Personally, I keep the intermediate answers, in my calculator, to full digits, then round at the end.
     
  4. Apr 19, 2017 #3
    Thank you. Yeah, overall, the textbook isn't the greatest, but I'm just following its lead for this course.
     
  5. Apr 19, 2017 #4

    haruspex

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    You were not asked for the time. You can get to the height directly by using the right SUVAT formula. Which one does not involve time?
     
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