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Yet another projectile motion

  1. Sep 11, 2007 #1
    1. The problem statement, all variables and given/known data
    A rifle has been sighted in for a 91.4m target. If the muzzle speed of the bullet is v0=427 m/s, what are the two possible angles theta1 and theta2 betweent the rifle barrel and the horizontal such that the bullet will hit the target? One of the angles is so large that it is never used in target shooting.


    2. Relevant equations
    I know you have to use trig identities, but I'm not sure how to get to that point.
    Maybe get the time it takes in one dimension to go half way?


    3. The attempt at a solution
     
    Last edited: Sep 11, 2007
  2. jcsd
  3. Sep 11, 2007 #2

    learningphysics

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    Write out the equations for vertical displacement and horizontal displacement.
     
  4. Sep 11, 2007 #3
    x-x0=delta x, y-y0=delta y, delta x=91.4m, delta y=0
     
  5. Sep 11, 2007 #4

    learningphysics

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    yes, write these using velocity, time, theta...
     
  6. Sep 11, 2007 #5
    is the v0 the same throughout the problem, or is it different for v0x, and voy?
     
  7. Sep 11, 2007 #6
    91.4m=vx*cos(theta)t, 0m=vy*sin(theta)t + 1/2(-9.80 m/s^2)t^2
     
  8. Sep 11, 2007 #7

    learningphysics

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    vy = v0sin(theta), and vx = v0cos(theta)

    The equations should be:

    91.4m=v0*cos(theta)t,
    0m=v0*sin(theta)t + 1/2(-9.80 m/s^2)t^2

    using these two equations try to solve for theta
     
  9. Sep 11, 2007 #8
    what do I do with the t variable? Is it the time is takes to go 91.4m at 427m/s? .214s.
     
  10. Sep 11, 2007 #9

    learningphysics

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    Solve for t in one equation... substitute into the other equation... then solve that equation for theta.

    Yes, it is the time it takes to reach 91.4m... but how did you get 0.214s ?
     
  11. Sep 11, 2007 #10
    from v0 in a one dimensional equation.
     
  12. Sep 11, 2007 #11

    learningphysics

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    the time here will be different because of the angle...

    use [tex]t = \frac{91.4}{v0*cos(theta)}[/tex]

    and substitute into the other equation...
     
  13. Sep 11, 2007 #12
    0m=(427m/s)*sin(theta)*((91.4m)/((427m/s)*cos(theta))+(1/2)*(-9.80m/s^2)*((91.4m)/((427m/s)*cos(theta))^2
    (0.225/cos^2(theta))=(91.4*sin(theta))/cos(theta)
     
  14. Sep 11, 2007 #13

    learningphysics

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    You can cancel cos(theta) from both sides... then use a trig. identity...

    but one thing I should have mentioned:

    0m=v0*sin(theta)t + 1/2(-9.80 m/s^2)t^2

    can be simplified to

    0m=v0*sin(theta) + 1/2(-9.80 m/s^2)t

    by dividing both sides by t... since we aren't dealing with the t = 0 case.

    So solving:

    91.4m=v0*cos(theta)t
    0m=v0*sin(theta) + 1/2(-9.80 m/s^2)t

    will also work...
     
  15. Sep 11, 2007 #14
    sin(2theta)=.00492, can't remember how to get theta from here.
     
  16. Sep 11, 2007 #15

    learningphysics

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    take the inverse sin of both sides... that will give 2theta = .... use your calculator to get the inverse sin of 0.00492. the calculator will only give one value... but there are two angles that give the same sin.

    Using 2theta = x (where x comes from the inverse sin), you solve and get theta = x/2


    You did everything correctly. But I just wanted to also show the formula when we don't plug in the numbers right away...

    R = v0*cos(theta)t (where R is the range)
    0 = v0sin(theta) - (1/2)gt

    0 = v0sin(thet) - (1/2)g[R/(v0cos(theta))]

    0 = 2v0^2sin(theta)cos(theta) - gR

    gR = v0^2 sin(2theta)

    [tex]sin(2\theta) = \frac{gR}{{v_0}^2}[/tex]

    plugging in R and v0 here I get 0.004913
     
    Last edited: Sep 11, 2007
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