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Finding time given with acceleration and height.

  1. Oct 31, 2014 #1
    1. The problem statement, all variables and given/known data
    An object is thrown from the ground into the air at an angle of 30 degrees to the horizontal. If this object reaches a maximum height of 5.75m, at What velocity was it thrown?

    a= -9.81 m/s
    h= 5.75 m
    I know that velocity is the one that I need to solve but I'm kinda finding it hard for myself to cope up how to solve for the time because I know that once I figure out the time, I would easily be able to solve for the velocity
    2. Relevant equations
    • a= Vf-vi/t
    • t=0-vi/a
    • t= -vi/a



    3. The attempt at a solution
    • a= Vf-vi/t
    • t=0-vi/a
    • t= -vi/a
    • T= -sin(30/-9.81
    • T=0.05s
    I know I probably used the wrong solution but i'm running out of option. I tried googling it but I can't find an answer that I could actually grasp in my mind. I have a quiz tomorrow and I'm afraid there might be a question like this and I might have a hard time figuring it out if I don't learn it by tonight. I hope you guys can help me. Sorry for the messed up format because I'm not really used to the settings here in forums :).
     
  2. jcsd
  3. Oct 31, 2014 #2

    Simon Bridge

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    ... hint: conservation of energy.
    BTW: what is sin(30deg)?
     
  4. Oct 31, 2014 #3

    haruspex

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    What happened to vi?
    There are 5 standard variables in the 5 SUVAT equations, each involving four of the variables.
    In any constant acceleration problem, identify the three variables whose value you know and a fourth whose value is to be determined. Pick the equation that involves those four.
    What other SUVAT equations do you know? (As Simon says, one of them corresponds to conservation of mechanical energy.)
     
  5. Oct 31, 2014 #4
    I'm not really sure. I had a similar question like this on my text book but instead of height is the given, it was the x displacement. My teacher wrote a solution of T=D/V > 8.59/v(cos)23
    then he substituted that for the time in D=Vit + .5at^2 which became D=vsin23(8.59/cos23) +0.5(9.81)(8.59/cos23)^2 which eventually gave the velocity for the horizontal. I just assumed that the 8.59/v(cos)23 to solve for time would have been similar to this question.
     
  6. Nov 1, 2014 #5

    Simon Bridge

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    Trying to copy what was done in class without understanding it may work but usually leads to error.
    Do you know what energy transformations are involved?
    Do you know how to separate the motion into vertical and horizontal components?
     
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