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Energy problem involving a projectile (tricky)

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data
    A ball (0.45kg) is kicked at 32m/s, 45° above the horizontal. If the average force of air resistance is
    1N, how far will it go?


    2. Relevant equations
    so I figure you would need energy to solve this question but I'm having trouble finding the total distance it travels, not just the horizontal distance but the parabolic-like distance. Am I on the right track? I have some of the things figured out but I'm finding it a little overwhelming.

    3. The attempt at a solution
    Loss to friction W=Fd W=(1)d
    Ep=mgh=mgdy=(0.45)(9.8)dy for potential energy at maximum

    Ek=1/2(0.45)(32)2 for kinetic energy at start

    Y direction
    Vo= 32 sin 45°= 22.63m/s
    V=?
    to find acceleration (this could be wrong) F=ma= 1=0.45a which means a=2.22 but the y component =1.57+g=11.37 down
    dy=?
    t=?

    X direction
    Vo= 22.63m/s
    V=?
    Did the same thing to find a here
    a=-1.57
    dx=?
    t=?
     
  2. jcsd
  3. Mar 25, 2014 #2
    It may be trickier than you think.
    I suppose the air resistance is in direction opposite to the velocity, so it changes direction.
    You will have variable accelerations on both x and y directions.
    What is the meaning of "average force" here? Is this form a book or you made it up?
     
  4. Mar 25, 2014 #3
    This was made by my teacher, as for the air resistance I think it means that at any point the force of friction with the ball against the air is 1N (of course it may be more or it may be less), just so we don't have to deal with increasing resistance as the ball speeds up.
     
  5. Mar 25, 2014 #4
    And what will be the direction of this force? This matters too.
     
  6. Mar 25, 2014 #5
    You're right that you don't have to deal with increasing resistance with speed but, as already pointed out, you do have to worry about the variable direction of the resistance.

    The only way I can think of solving this is using a numerical method.
     
  7. Mar 25, 2014 #6
    But could you not just completely overlook that and just use the work formula W=F*d and since F is 1N we know that the amount of energy lost to friction (air resistance) is equal to the distance? And solve using conservation of energy instead, or am I missing something important?
     
  8. Mar 25, 2014 #7

    NascentOxygen

    User Avatar

    Staff: Mentor

    But you don't know the distance it travels, nor the time spent in the air. The path taken is roughly parabolic, not exactly parabolic.
     
  9. Mar 25, 2014 #8
    Exactly my problem, if there were no air resistance, one could probably use a quadratic equation to model its flight path. Any advice as to how I might tackle it?
     
  10. Mar 25, 2014 #9
    Numerically.
     
  11. Mar 25, 2014 #10
    Last edited: Mar 25, 2014
  12. Mar 26, 2014 #11
    This would be pretty hard to solve using work and energy, considering you'd have to find the arc length of the trajectory (which isn't always an easy task) to know the distance over which your air resistance is applied.

    Use Newton's 3rd law and the Kinematics, it'll be a lot easier.

    You can see that $$a_{x} = -\frac {F_{D}}{m}$$ and $$a_{y} = - ( g + \frac {F_{D}}{m} )$$

    These are found by comparing forces in each direction. Note ##F_{D}## is the drag force/air resistance.

    So, plugging into the kinematics for the y direction we find $$t = \frac {v_{0}\sin \theta}{g+\frac {F_{D}}{m}}$$

    Now we should multiply this time by 2 considering it is the time to get to the top of the trajectory, but we want the total time.

    So, plugging into another one of the kinematics for the x direction: $$x = v_{0}\cos \theta t - \frac {F_{D}}{m}t^2$$

    and using ##t = 2\frac {v_{0}\sin \theta}{g+\frac {F_{D}}{m}}##

    You can simplify all of this into a decently clean equation for the total trajectory.

    $$X_{traj} = \frac {v_{0}^2}{g+\frac {F_{D}}{m}} (\sin 2\theta - \frac {F_{D} - \cos2\theta}{mg+F_{D}})$$
     
  13. Mar 26, 2014 #12
    The above formula would be correct if F_D was the same in the x-direction and the y-direction. However, the problem strictly as interpreted gives the average value or magnitude as constant without specifying the direction. This makes the problem difficult to solve if not impossible using a direct method.

    Perhaps the OP can clarify if the drag force F_D is supposed to be the same in each direction or if it is a constant magnitude whose direction is opposite to the velocity?
     
  14. Mar 26, 2014 #13
    That's a cute little program. The drag force it is using is almost certainly one that is proportional to velocity. The difference with your problem is that the magnitude in your problem is given as a constant. You might want to clarify this with your teacher.
     
  15. Mar 26, 2014 #14
    This is a rather intuitive principle of a consistent medium like air, and is a fair assumption that the force would have the same resisting value for both components of motion.

    Also, for the program, if you leave the radius and drag force alone and merely enter in the values explicitly given in the question, there's roughly a 0.2% error between the experimental and calculated trajectory.

    You're right though. This question is a bit vague on many values, considering drag force is not really an easy value to calculate.
     
  16. Mar 26, 2014 #15
    So even if there is no motion along some direction, there will be a drag force along that direction?
    It does not seem very intuitive.
    If you drop a ball, will it experience a lateral drag force?
    Or if you launch it at 89 degrees upward, the drag force for the horizontal component (which is almost zero) will be the same as for the vertical motion?
     
  17. Mar 26, 2014 #16
    Considering it is given purely as a constant force, then yes. In this equation, that's exactly what it is. If your drag force is a constant and there are no other forces in the horizontal component, your acceleration will always be $$a_{x} = -\frac {F_{D}}{m}$$

    And summing up your forces in the y direction will give: $$ma_{y} = -mg - F_{D}$$
    I will say you are right with some of the extreme cases. The equation I gave was under the assumption that the projectile stayed in motion for the entire time it was in the air. If the projectile's horizontal component is so small that the air resistance stops it before it even hits the ground, then you would calculate your trajectory using a time value of $$t = \frac {mv_{0}\cos \theta}{F_{D}}$$
     
  18. Mar 26, 2014 #17
    No, not even in the (confusing) text of the problem there is anything to grant the assumption that each component of the force has a constant value of 1N.
    How do you calculate the "average" to get 1N with your assumptions of 1N on each direction?

    And this is not considering the fact that the assumptions are unnatural.
    At the top part of the trajectory the vertical motion is much slower than the horizontal. However you assume that the drag is the same.

    I think is more "natural" to question the validity of the problem rather than trying to force reality to fit the problem.
     
  19. Mar 26, 2014 #18
    The problem is unrealistic, but the equation fits the problem. If you somehow had a drag force that was constant (which is retarded, considering it's a function of the velocity normally) then you could use that equation and get the correct answer. So yes, obviously that equation wouldn't even work in the real world. We'd probably set it up as a first order linear differential equation.
     
  20. Mar 26, 2014 #19
    What kind of averaging (in your model) will give you an average drag force of 1N, in order to satisfy the problem?
    Your total drag force is about 1.4 N all the time.
     
  21. Mar 26, 2014 #20
    I am afraid, I know only as much as is written in my original post, as I copied the question out exactly as it was written. It probably should be noted, that my teacher came up with the question which means he could have possibly made an error in his calculations, perhaps assuming it was a parabola, as unlikely as that would be for a teacher to make a simple error like that.
     
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