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Homework Help: General Basic Projectile Motion Question

  1. Aug 9, 2013 #1
    In projectile motion problems, where an object is thrown (not horizontally) off of a cliff, does only the cliff's height matter to this equation:

    [itex] y = v_y * (t) + (a * t^2)/2 [/itex]

    What I'm confused about is the little bit of height gained by the object when it is first thrown off of the cliff. Is this height insignificant to the overall problem?
  2. jcsd
  3. Aug 9, 2013 #2
    Here y is displacement in the vertical direction, what is the displacement of the particle? Does it depend on the maximum height reached?.
  4. Aug 9, 2013 #3


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    Staff: Mentor

    If it will influence the significant figures of any results, then by definition it is significant :smile:
  5. Aug 9, 2013 #4
    In projectile problems, you have a set of parametric equations where the vertical height is a function of t, and the horizontal height is a function of t. The equation you show applies to the vertical height and is really not complete in that the initial distance, or height in this case, is not in the equation. You question "the little bit of height". Actually there is normally a lot of height depending on the angle that you throw or shoot the projectile. The equation you show gives the height as a function of time, t, above the elevation of the cliff. If you want an equation that shows the height of the object above the ground it would be
    y = d0 + vo(t) + (1/2)at^2 where d0 is the height of the cliff.
  6. Aug 9, 2013 #5

    Oh! so since we're talking about displacement, only the starting and ending positions of the particle matter? It doesn't matter where the particle goes in between 310 m and 0 m.

    Thank you! (unless I got this wrong...again!)
  7. Aug 9, 2013 #6
    I think what you all are saying is that this height is indeed significant (please ignore if that's not what you actually intended to convey!), which is what I thought when I was first trying to solve this problem:

    A student decides to throw a rock off a cliff. The rock leaves the top of the cliff at a speed of 35 m/s and at an angle of 50 degrees above the horizontal. If the cliff is 310m tall, how far down range is the fruit when it hits the ground? Neglect air resistance.

    This was a worked problem, and I agreed with all of the work, except that:
    y = v*t +(1/2) a * t^2
    -310m = (26.8 m/s) * t + (1/2) * (-9.8 m/s^2) * t^2

    And I thought y should have had the extra bit of height gained when the rock was thrown.

    As explained by the first poster, is y change in displacement, not greatest vertical distance rock travels?
  8. Aug 10, 2013 #7


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    Gold Member

    Did you miss a negative there? Is that the full working to the question? When the rock is projected upwards from the height 310m, there will come a time when it reaches the same height of 310m as it falls back down to earth. If that isn't the full working to the question, then that eqn will give the time taken for the rock to fall to the ground when it reaches that height of 310m again.
  9. Aug 10, 2013 #8
    Exactly. It seems a bit paradoxical at first.
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