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PLEASE HELP - Max. Height of a Projectile

  1. Feb 26, 2005 #1
    I've been working on this problem for almost TWO HOURS ( :yuck: ) and I still don't understand it. Please help!

    We're in the Work, Power and Energy unit right now and I have to solve this problem:

    A 3.18-kg ball is launched from a cliff at 60 degrees from the horizontal. Determine the maximum height (in meters to two decimal places) that the ball reaches above the valley floor if it has an initial speed of 41.6 m/s. The height of the cliff is 37 meters. Ignore air resistance.

    First of all, this is part of a problem set on Potential and Kinetic Energy, so I have to use only Work and Energy concepts to solve it.

    I first assumed that the kinetic energy at the ball's maximum height was instantaneously zero (I looked this up online and found many web sites claiming just that fact) and solved for the height that way:

    PEi + KEi = PEf
    (3.18kg * 37 m * 9.8m/s^2) + (1/2 * 3.18kg * (41.6m/s)^2) = (9.8m/s^2 * 3.18kg)height
    3904.6584 J = 31.164h
    h = 125.2938776 meters

    I entered that into our class database and it was wrong.

    It occurred to me that perhaps the kinetic energy at a projectile's maximum height isn't zero. I recalled what we had learned in the Vectors and Projectiles unit and remembered that at the maximum height, the vertical velocity would be zero but the horizontal velocity would be vi * cos theta.

    I went to my friend google and searched for projectile problem solving methods. Sure enough, I found the same explanation on numerous physics sites: at its maximum height, a projectile's horizontal velocity is vi * cos theta.

    So, I tried solving the problem that way, adding in the value for KEf:

    PEi + KEi = PEf + KEf
    (3.18kg * 37 m * 9.8m/s^2) + (1/2 * 3.18kg * (41.6m/s)^2) = (9.8m/s^2 * 3.18kg)height + (1/2 * 3.18kg * (41.6sin60)^2)
    3855.0504 J = 31.164h
    height = 123.7020408

    Still wrong.

    Well, at that point I was really frustrated because I was convinced that I was doing the problem correctly. I went to google and searched for different ways to solve projectile max height problems.

    I found a .pdf file from some university concerning problems of the same type. It offered this explanation:

    ...use conservation of energy to relate the horizontal and vertical components of the velocity and the points of launch to the maximum height. If you then note that in the horizontal direction there is no force, and thus no acceleration, ie the horizontal component of the velocity does not change, you should find an expression for the maximum height:

    height = (vi^2 * sintheta^2)/(2g)

    Well, that would be the same as max height = vertical velocity component/2g.

    I used that equation and ended up with 66.22040816 m. I added this to the initial height of 37 m and it turned out to be correct.

    I got the correct answer and I'm happy, but the trouble is that we have a unit test on tuesday and I don't understand why that equation worked, where it comes from, or how it relates to potential and kinetic energy and not just vectors and projectiles. Also, I'm almost certain we're going to have the exact same problem (with different values, of course) on the test and if I just write that equation as my work, my physics teacher is going to wonder where I got it from.

    I'd really appreciate the help if anyone could offer any explanations or different ways to solve this problem. Thanks!
     
    Last edited: Feb 26, 2005
  2. jcsd
  3. Feb 26, 2005 #2

    dextercioby

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    Homework Helper

    You needn't worry about the horizontal (always nonzero during the entire flight) component of the velocity,but only about the VERTICAL one.In this case,it's basically like a simple problem in 1D motion in constant gravity field...Apply the law of conservation of mechanical energy,putting for "v" in the square (from the KE term) as v*sin theta (the vertical component).

    Daniel.
     
  4. Feb 26, 2005 #3
    I thought that the vertical velocity at maximum height would be zero, though...

    Anyway, I tried using v*sin theta as the v at KEf and I got the wrong answer:

    PEi + KEi = PEf + KEf
    (3.18kg * 37 m * 9.8m/s^2) + (1/2 * 3.18kg * (41.6m/s)^2) = (9.8m/s^2 * 3.18kg)height + (1/2 * 3.18kg * (41.6sin60)^2)
    3855.0504 J = 31.164h
    height = 123.7020408

    I only got the right answer when I used height = (vi^2 * sintheta^2)/(2g). I'm confused as to why the second equation works but the first one doesn't.

    Ack.
     
  5. Feb 26, 2005 #4

    dextercioby

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    Coose the 0/origin for the PE in the point of the launch...

    Daniel.
     
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