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Kinematics Min/Max angles of a projectile

  1. Jun 21, 2013 #1
    1. The problem statement, all variables and given/known data
    A football player attempts a 30m field goal. If he is able to impart a velocity u of 30 m/s to the ball,compute the minimum angle θ for which the ball will clear the crossbar of the goal. (Hint: Let m = tanθ .)
    hw.png


    2. Relevant equations
    V0x = V cosθ
    V0y = V sinθ
    We get t as a function of θ , using : X=X0 + V0xt => t=1/cosθ
    And from the Y direction using Y = Y0 + V0yt + -g/2t^2
    and then we substitute t from the x equation , in the y equation.



    3. The attempt at a solution
    Now , i substituted the both equation and after hours of trying and simplifying , i came up with this equation :
    30sinθcosθ - (cosθ)^2 - 4.905 = 0
    and i got stuck on how should we get 2 values of θ from this equation.
    I really need help , i got the idea of the problem but i started giving up :(
     
  2. jcsd
  3. Jun 21, 2013 #2

    tiny-tim

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    welcome to pf!

    hi ieaglei! welcome to pf! :smile:

    hmm :rolleyes: … ​
    :wink:
     
  4. Jun 21, 2013 #3
    You might consider this cheating but you could use a graphing calculator to solve.

    Sure this is correct? - (cosθ)^2
     
    Last edited: Jun 21, 2013
  5. Jun 21, 2013 #4
    I saw a solution for a very similar question , and they got an equation similar to my equation above , and then they wrote : Using trail and error we get : θ1 and θ2
    and now I don't want to waste my whole life putting numbers in the calculator to find the angles , i want the right way to solve this :(
    and I already tried plotting the graph , and what i got is a sinusoid with infinity solutions which made me more confused .
    And by the way , i know the final answer which is θ mini = 15.43 degree
     
  6. Jun 21, 2013 #5
    I don't think you entered the 3m crossbar heighth correctly in your equation. You used 1m.
    I graphed the solution on my TI84 and got the correct answer.
     
  7. Jun 21, 2013 #6
    Man , I saw alot of problems on this site similar to mine , and I also saw your reply's :X
    and i had no benefits of them , I want some serious hints , i want to go on and solve it :(.
     
  8. Jun 21, 2013 #7
    First, get the correct equation to solve.
    Then learn how to solve graphically if not analytically.
     
  9. Jun 21, 2013 #8
    Oh , sorry the real equation i got was : 30sinθcosθ - 3(cosθ)^2 - 4.905 = 0
    not : 30sinθcosθ - (cosθ)^2 - 4.905 = 0
    Can you explain how you got the right answer?
     
  10. Jun 21, 2013 #9
    I cheated and used a graphing calculator.
    So far, I have not found an analytic solution.
     
  11. Jun 21, 2013 #10
    I have the same TI-84 plus silver edition as a program on my pc , but I don't know how to use it , can you just tell me how to plot that equation :( , I know I'm asking so much .. but this is the last :)
     
  12. Jun 21, 2013 #11
    Its easy. Plot y = 30sinθcosθ - 3(cosθ)^2 - 4.905 then look for the zero crossing. On a TI-84, once you see the graph cross zero, do (2nd)(trace)(2) to find the zero crossing. Put the cursor to the left of the crossing, then to the right of the corssing, hit enter and look at the answer.
     
  13. Jun 21, 2013 #12

    haruspex

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    You were given the hint m = tanθ. Easiest is to divide through by cos2θ and then turn all the trig references into functions of m. you should get a quadratic.
     
  14. Jun 21, 2013 #13
    Hmmmm... seems like if you divide by (cos theta)^2 you would have.
    30 tan(theta) -3 -4.905/(cos(theta))^2 = 0

    so, 1/(cos(theta)^2 = sec^2(theta)=1-tan^2(theta)

    Ah Ha!
     
    Last edited: Jun 21, 2013
  15. Jun 21, 2013 #14

    haruspex

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    sec^2(theta)=1+tan^2(theta)
     
  16. Jun 22, 2013 #15
    Thank you guys .. with your help , i reached the answer .
    As you said : making the equation in terms of tanθ..
    we will get : ( 4.905tan^2(θ) - 30tan(θ) + 7.905 = 0 )
    replacing tanθ with x , we get a quadratic equation .. solving it we get 2 answers ..
    tan^-1 (x1) = θ1
    tan^-1 (x2) = θ2

    problem solved and got the correct answer.
    thread can be closed <3 thanks all
     
  17. Jun 22, 2013 #16
    But i got theta is equal to 5.84 and 0.27 and when i subsitute in y equation to check my answer it gives me -1.8
     
  18. Jun 22, 2013 #17
    oooh i got m..forget to take tan^-1...Thanks alot for your help :)
     
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