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Projectile motion cannon

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data
    The Great Projecto is fired from a cannon at a velocity of 29.8 m/s at an angle of 50.4 degrees. He lands in a net which is 2.0 m above his starting elevation. What is his velocity when he hits the net?

    2. Relevant equations
    sin=opp/hyp
    cos=adj/hyp
    3. The attempt at a solution
    Using trigonometry I found that:
    vy1: 22.96 m/s
    vx1: 18.995 m/s
    vx2: 18.995 m/s
    And I know that
    ay(acceleration of y)= -9.8m/s^2
    Since the question tell us that the net is 2.0 m above the starting elevation, dy+2= the height of the net...but how will I find the starting elevation? I don't know where to start.
     
  2. jcsd
  3. Dec 4, 2016 #2

    kuruman

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    You cannot solve this problem using just trigonometry. You have to use the kinematics equations for projectile motion.
     
  4. Dec 4, 2016 #3
    But none of the equations let me find the displacement without having another piece of information?
    v2^2=v1^2+2ad>>>>I don't have d
    v2=v1+at>>>>>I don't have t
    d=v1^2+1/2a(t)^2>>>>I don't have d or t
    d=v2^2-1/2a(t)^2>>>>I don't have d t or v2
    d=(v1+v2/2)t>>>>I dont have d or t
     
  5. Dec 4, 2016 #4

    PeroK

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    @gungo what about using the velocity components in the two directions?
     
  6. Dec 4, 2016 #5

    kuruman

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    You know the displacement and the initial velocity in the vertical direction. Use them to find the time of flight first. Having the time of flight, you cn find anything else you need.
     
  7. Dec 4, 2016 #6

    PeroK

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    You don't need the time of flight.
     
  8. Dec 4, 2016 #7

    kuruman

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    That's right, you don't. You still need a kinematic equation in the vertical direction.
     
  9. Dec 4, 2016 #8
    Got it, thanks!
     
  10. Dec 5, 2016 #9

    PeroK

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    Note that ##v^2 =u^2-2gh## works for projectile motion, where ##h## is the upwards displacement in the ##y## direction.

    You can derive this by:

    ##v^2 = v_x^2 + v_y^2 = u_x^2 + u_y^2 -2gh = u^2 -2gh##

    Using the constant velocity in the x direction and the acceleration of gravity in the y direction.

    Note also that you didn't need the angle of projection. When you come to learn about energy you'll see why.
     
  11. Dec 5, 2016 #10

    kuruman

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    The problem is asking for the velocity, which means two components. Although you don't need the projection angle to find the final speed as @PeroK pointed out, you still need the angle to find how that speed is divided between the x and y directions.
     
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