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When to use certain kinematic equations

lamborghiniMD

Full Member
Jun 25, 2012
136
8
  1. Pre-Medical
    Okay, so here's a problem from a practice test:

    "Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in a time, t. If the value of g where reduced to g/6, then t would:

    a) incr by 6
    b) incr by 6^.5
    c) decr by 6
    d decr by 6^.5

    Okay, so I narrowed it down to a & b, and I WAS going to pick a (the correct answer), but then of course I had to doubt myself and pick b.

    I chose a at first using t = v/g. but then I switch to b, because I used h = .5at^2.

    Now I'm confused. Both equations are valid for time, but why was the t = v/g correct for this problem?
     

    cloak25

    Full Member
    Feb 28, 2012
    116
    0
      Okay, so here's a problem from a practice test:

      "Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in a time, t. If the value of g where reduced to g/6, then t would:

      a) incr by 6
      b) incr by 6^.5
      c) decr by 6
      d decr by 6^.5

      Okay, so I narrowed it down to a & b, and I WAS going to pick a (the correct answer), but then of course I had to doubt myself and pick b.

      I chose a at first using t = v/g. but then I switch to b, because I used h = .5at^2.

      Now I'm confused. Both equations are valid for time, but why was the t = v/g correct for this problem?

      I think you use h = .5at^2 when you assume initial velocity is 0. In this case, the ball is thrown up so it has an initial velocity, therefore you cannot use h=0.5at^2.

      Instead you can use d= vot + 0.5at^2 where d = 0 because the ball returns to its original height (0 displacement). So 0 = vot + 0.5at^2

      since acceleration is negative, vot = 0.5at^2. time cancels and you get t = 2vo/g, which is the equation they used to get the answer.
       
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      flashlightpen

      Full Member
      Jan 27, 2012
      20
      2
      1. Pre-Medical
        Okay, so here's a problem from a practice test:

        "Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in a time, t. If the value of g where reduced to g/6, then t would:

        a) incr by 6
        b) incr by 6^.5
        c) decr by 6
        d decr by 6^.5

        Okay, so I narrowed it down to a & b, and I WAS going to pick a (the correct answer), but then of course I had to doubt myself and pick b.

        I chose a at first using t = v/g. but then I switch to b, because I used h = .5at^2.

        Now I'm confused. Both equations are valid for time, but why was the t = v/g correct for this problem?

        Hey man!

        Total off topic, but a pretty cool tip I got :D Whenever a question mentions time, use kinematics. If they dont want time, just use energy. Sounds kind of simple, but its been really helpful for me :thumbup:
         
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