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Finding the initial velocity and time period of a projectile

  1. Feb 25, 2015 #1
    1. The problem statement, all variables and given/known data

    In the Fig.
    What should be the initial velocity (magnitude and direction) of the arrow? Also calculate the time of flight.
    10996590_972978676065348_864850896699954934_n.jpg 10996590_972978676065348_864850896699954934_n.jpg
    2. Relevant equations


    3. The attempt at a solution
    I have formulated 2 equations:
    v*cos(theta) * t = 20
    and 3 = (v * Sin(theta) * t) + (9.81 * t^2)/2
    when t = time of flight
    and v = initial velocity

    I have further gone to a final equation
    v = (2000+2000(tan(theta))^2)/ (3 + 20 Tan(theta))
    Any help please.
     
  2. jcsd
  3. Feb 25, 2015 #2

    PeroK

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    Do you think there is only one way to hit the apple?
     
  4. Feb 25, 2015 #3
    Well, intuitively speaking, I think that this problem will have a range of values as we can hit the apple along a variety of trajectories.
     
  5. Feb 25, 2015 #4

    PeroK

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    That's correct. So, could you clarify what you are trying to calculate? For example, there will be a minimum value of v for which you can hit the apple.
     
  6. Feb 25, 2015 #5
    But won't there be an optimal solution to this, the solution that takes the least time? I mean the value of v that takes the least time?
     
  7. Feb 25, 2015 #6

    PeroK

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    There's not really going to be a least time, as the least time will be for the fastest v.

    I think there would be a longest time, associated with the lowest possible velocity.
     
  8. Feb 25, 2015 #7
    Oh yes! How stupid of me to ask :P But how will we find the lowest value of v?
     
  9. Feb 25, 2015 #8

    PeroK

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    Are you sure you are not missing a constraint from the problem? It's quite tricky to calculate the lowest v to hit a point above the ground.
     
  10. Feb 25, 2015 #9
    well the problem, unfortunately, in front of me only gives this information. I think that there will be different minimum velocity for every possible angle. Am I right? and how do we find the range of angles possible?
     
  11. Feb 25, 2015 #10

    PeroK

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    I'm pretty sure it's not the problem you're supposed to be doing, but if you want a challenge.

    First step is to express the trajectory as a (parabolic) function. Here's the formula. Try to derive this:

    ##y = xtan\theta - \frac{gx^2}{2v^2}sec^2\theta##
     
  12. Feb 25, 2015 #11
    I have already done that:
    I formulated two equations:
    1. vCosθ * t = 20
    as x =20
    2. 3 = vSinθ(t) + g(t^2)/2
    as y =3 and then substituting the value of 't' i.e. 20/vcosθ in the second equation
     
  13. Feb 25, 2015 #12

    PeroK

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    Okay, but remember g = -9.81 in your equations.

    If you're trying to minimise v, then you need to maximise ##\theta## which is the same as maximising ##tan\theta##. Note that ##sec^2\theta = tan^2\theta + 1##

    So, aim for a quadratic equation in ##tan\theta##.
     
  14. Feb 25, 2015 #13
    Okay, that's what I have reached at:

    is the answer 89.7 degrees?
     
  15. Feb 25, 2015 #14
    And also could you tell me the intuitive reason why we took g as -9.81( I am sometimes confused by this). I do get why we took this value in this particular problem as we had to get a maximum value of tanθ hence we get a parabola with a maximum point.

    And, am I right in thinking that there will be a different minimum velocity for every possible angle?
     
  16. Feb 25, 2015 #15

    PeroK

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    That can't be correct.
     
  17. Feb 25, 2015 #16
    y = xtanθ - (gx^2 secθ^2)/2 Comment: as we take g =- 9.81 so negative sign)
    2y = 2xtanθ-gx^2secθ^2
    2y = 2xtanθ-gx^2-gx^2(tanθ)^2
    2y+gx^2=2xtanθ-gx^2(tanθ)^2
    now plugging in the values: x = 20; y =3
    2(3) +(9.81*400)=2(20)tanθ-(9.81*400)(tanθ)^2
    now by substituting any variable say 'x' as tanθ
    -3924x^2 + 40 x - 3930 = 0
    Then differentiating it and putting that to zero and getting the value of x to be 196.2 and then using tanθ = 196.2 and then finding out the value of θ.
     
  18. Feb 25, 2015 #17

    PeroK

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    I might have led you astray! We need to maximise the quadratic function for a given v. And, if the maximum value is y, we have the minimum value for v. It's not maximising theta. We need to maximise:

    ##f(\theta) = xtan\theta - \frac{gx^2}{2v^2}(1 + tan^2\theta)##

    I told you it was tricky!
     
  19. Feb 25, 2015 #18
    But is not the 'y'(vertical displacement) always constant as the displacement will always be 3 no matter what? I mean the final displacement will be 3. Thanks for the help :D
     
  20. Feb 25, 2015 #19

    PeroK

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    That's not the point. Suppose we fire the arrow at a velocity v. That function gives the height at x for a given ##\theta##. So, first we calculate the maximum height at x for a given v (for any ##\theta##).

    If that maximum height happens to be y (=3 in this case), then we have found the minumum v. Think about it. If we could hit (x, y) with a lower velocity than v, then we could hit a higher point (x, something higher than y) at velocity v.

    In other words, at the minimum v, we can only just hit the target. And no higher.
     
    Last edited: Feb 25, 2015
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