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Homework Help based on the equations of motion

  1. Sep 23, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle is projected vertically upwards at 30 m/s. Calculate (a) how long it takes to reach its maximum height, (b) the two times at which it is 40 m above the point of projection, (c) the two times at which it is moving at 15 m/s.


    2. Relevant equations[/b
    final velocity (v)
    initial velocity (u)
    acceleration due to gravity (a) = 10m/s^2
    time (t)
    Distance or displacement (s)
    v^2=u^2 + 2as
    v=u+at
    s=ut+1/2 at^2
    s=1/2(u+v)t


    3. The attempt at a solution
    I first find the maximum height, this is where i used the equation v^2=u^2 +2as
    0^2=30^2 + 2(10)s
    s=45 m maximum height

    then i used that value and plugged into the equation to find time
    45= 1/2 (30+0)t
    t= 3seconds

    therefore total time is 3 seconds for the particle to reach its maximum height.

    part (b) is where i am stuck

    however i move on to part (c)
    and i plugged in the values in the equation
    15=30-10t
    t=1.5 seconds

    however the question asked for two (2) times. so i am also stuck with finding the second time...
     
  2. jcsd
  3. Sep 23, 2013 #2

    berkeman

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    Staff: Mentor



    Welcome to the PF.

    When they are asking for 2 times, one will be on the way up, and the other on the way down. There is a squared term in the equation for the vertical motion, and you should be able to use both roots in solving for the two times. Does that help?
     
  4. Sep 23, 2013 #3
    no sorry im lost....
     
  5. Sep 23, 2013 #4

    gneill

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    Staff: Mentor

    When the problem says, "...it is moving at 15 m/s", this is giving you a value for the speed of the projectile.

    A speed of 15 m/s can mean the projectile is going up (+15 m/s) or down (-15 m/s). Speeds don't have directions associated with them, velocities do :wink:
     
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