Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Projectile motion without time and initial velocity

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data

    A shotputter projects the shot at 42 degrees to the horizontal from a height of 2.1 m. It lands 17 m away horizontally. Next, he gives it the same initial speed but changes the angle to 40 degrees. What effect does this have on the horizontal range?

    2. Relevant equations



    3. The attempt at a solution

    I tried solving this by algebraic means but failed to come to the answer of 0.02m less which is given in the back of the book. Not having initial velocity or time makes this problem difficult to solve but i know that by understanding how to solve this problem I can understand how to do similar problems. I would greatly appreciate any assistance. I like to do things my self so I am not asking you to do the work for me, but perhaps send me in the right direction.
     
  2. jcsd
  3. Jun 5, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    What did you try?
    Hint: Use the initial data to solve for the initial velocity.
     
  4. Jun 5, 2010 #3
    So the given data is as follows:

    angle = 42 degrees
    yf = 0m
    yi = 2.1m
    xi = 0m
    xf = 17m

    I used the given data to produce the following six equations where "t" is time and "V" is velocity and "g" is the acceleration due to gravity -9.8m/s^2.

    1) Vfx = (Vi)(cos(angle))
    2) xf = (Vi)(cos(angle))(t)
    3) Vfy = (Vi)(sin(angle)) + (g)(t)
    4) (Vfy)^2 = ((Vi)(sin(angle)))^2 + 2(g)(-yi)
    5) (-yi) = (1/2)((Vi)(sin(angle)) + Vfy)(t)
    6) (-yi) = (Vi)(sin(angle))(t) + (1/2)(g)(t^2)

    I solved equation (2) for time.

    2) t = (xf)/((Vi)(cos(angle)))

    I substituted for time in equation (6) and solved for initial velocity yielded the wonderfully long equation (7).

    7) Vi = sqrt( ((g)(xf^2)) / ((2)((-yi) - (xf)(tan(angle))))

    I substituted the data and yielded the following answer.

    Vi = 9.019586627 m/s

    I adjusted the angle and used equation (4) to solve for final vertical velocity and yielded:

    Vfy = -8.647142668 m/s

    I substituted the value for final vertical velocity into equation (3). I used trigonometry and the new angle and calculated initial velocity to solve for initial vertical velocity and substituted that into equation (3) as well and solved for time and yielded:

    t = 1.473961346 s

    I double checked the time using equation (6) and finding the roots of the parabolic function using the quadratic formula. Next I used equation (2) to find the final horizontal position and I yielded:

    xf = 10.18419474 m

    An obviously drastic difference compared to the answer given in the book of 0.02 m less. I even thought perhaps I misunderstood the question and I found the range for both angles using the calculated initial velocity and the following equation.

    R = ((Vi^2)(sin((2)(angle)))) / (g)

    The difference I calculated was about 0.08 m.

    So that is all my work with the exception of the little scribbles and such. Please if you have the time, look over it and either show me where I went wrong or tell me the book is wrong. I appreciate any help. Thank you Doc Al for the help so far :D.
     
  5. Jun 5, 2010 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Careful with signs. The acceleration is -g, not g.

    Redo your calculation for the initial velocity.
     
  6. Jun 6, 2010 #5
    Notice that my g = -9.8 m/s^2 . So the negative sign makes up for the presence of a positive g rather than a negative. I will redo my calculations now and get back to you.
     
  7. Jun 6, 2010 #6
    Ok, I believe i found my source of error. When solving for initial velocity equation (7) should turn out as following:

    7) Vi = sqrt( ((g)(xf^2)) / ((2)(cos^2(angle))((-yi) - (xf)(tan(angle))))

    I left out the cos^2(angle). I redid the calculations and yielded the answer in the back of the book. Thank you for your help. I should be more meticulous with my algebra.
     
  8. Jun 6, 2010 #7

    Doc Al

    User Avatar

    Staff: Mentor

    Excellent. I did the calculation myself, so I knew you had an error in there somewhere.
     
  9. Jun 9, 2010 #8
    You're not looking at this problem in the right way ... you don't need any formulas to answer the question.

    Hint: For any given projectile, what angle gives you the maximum range?
    What happens to the range as you increase that angle?
    What happens to the range as you decrease that angle?
     
  10. Jun 9, 2010 #9

    Doc Al

    User Avatar

    Staff: Mentor

    What do your questions have to do with this particular problem? And how you suggest getting a numerical answer, which is what is asked for, without the use of 'formulas'?
     
  11. Jun 9, 2010 #10
    The original question doesn't specify a numerical answer. The question is: What is the effect on the horizontal range when the trajectory angle is changed from 42 degrees to 40 degrees.

    The answer is: The horizontal range is reduced.


    My questions apply to this problem as such:

    Q1: For any given projectile, what angle gives you the maximum range?
    A1: The answer is 45 degrees.

    Q2: What happens to the range as you increase that angle?
    A2: The range reduces as the angle is increased.

    Q3: What happens to the range as you decrease that angle?
    A3: The range reduces as the angle is decreased


    So, the farther your trajectory deviates from 45 degrees, the lower your range will be.
     
  12. Jun 9, 2010 #11

    Doc Al

    User Avatar

    Staff: Mentor

    Read the entire first post (and the rest of the thread, for that matter) and you'll see that a numerical answer is required.

    That's true when the initial and final heights are the same, but not necessarily when they are not.
     
  13. Jun 10, 2010 #12
     
  14. Jun 11, 2010 #13
    You're absolutely correct. It turns out that the optimum angle of trajectory is approximately 41.47943509 degrees, which gives a horizontal distance of about 17.00223123 meters (at the same initial velocity)
     
  15. Jun 11, 2010 #14
    I still contend that the question, as stated, does NOT require a numerical answer, even though the answer in "the book" states that the horizontal distance is 0.02 m less. In this case the numerical part of the book's answer is additional information.

    Suppose the book's answer was: The horizontal distance was reduced by 0.02 m and the sun is yellow. This would not mean that a numerical answer and the color of Earth's main source of heat were required ... it's just extra information.

    The answer to the question, as stated, remains: "The horizontal distance is reduced."
    THAT is the effect of changing the trajectory from 42 degrees to 40 degrees.

    For the question to REQUIRE a numerical answer, it should have been stated something like:
    "What is the change in the horizontal distance due to the change in trajectory?"
     
  16. Jun 11, 2010 #15
    Just matter of perspective, but I feel that this is a more correct description:
    g IS the acceleration (not -g). But, since we consider downward acceleration to be negative, the value of g is -9.8 m/s/s.
     
  17. Jun 11, 2010 #16
    The source of your mis-calculation stems from the fact that you show Vi (the initial velocity) to be 9.019586627 m/s when, in fact, this is the value of Viy (the initial vertical velocity).

    Try starting with an initial velocity of 12.13705097 m/s and find the final X position at a trajectory of 42 degrees.
     
  18. Jun 11, 2010 #17

    Doc Al

    User Avatar

    Staff: Mentor

    The point of my comment was that in common usage, g represents the magnitude of the acceleration due to gravity, thus g = 9.8 m/s2. The sign depends on the convention used; in this case a = -g is appropriate. (But that wasn't the issue here, as the OP used g = -9.8 m/s2.) The important thing is that a = -9.8 m/s2.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook