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Projectile Motion Cannon Ball Question

  1. Oct 4, 2012 #1
    1. The problem statement, all variables and given/known data
    A cannon shot a ball at 90° and reached 2.19m high. From this I need to find Vi of that and find time, Vx, Vy, Dx, and Dy of a projectile being shot from this cannon at 35.00°.
    I would just like to confirm I have done this correctly.
    Dy=2.19m
    θ=90.00°

    2. Relevant equations
    Vf2-Vi2/2g
    Vx=Vsinθ
    Vy=Vcosθ
    Tt=2(Vsinθ)/g
    Vy=Vi+at
    Dx=Vxt
    Dy=Vit+at2/2

    3. The attempt at a solution
    2.19= 02-vi2/-19.6
    Vi=6.552m/s

    Tt=2(6.552sin35)/9.8
    Tt= 0.7670s

    Vx=3.758m/s
    Vy= 5.367m/s

    Dx= Vxt
    Dx= 3.758(0.7670)
    Dx=2.882m

    To find height at 1/2 time:
    d=Vit+at2/2
    d=3.758(0.3835)+(-9.8x0.38352)/2
    d=0.7205

    I concluded with:
    Tt= 0.7670s
    Max height= 0.7205m
    Range= 2.882m

    Is this correct?
     
  2. jcsd
  3. Oct 4, 2012 #2
    For something shot upward at an angle of 35 degrees how can the following be true? Vy > Vx ???

    Vx=3.758m/s
    Vy= 5.367m/s
     
  4. Oct 4, 2012 #3

    tiny-tim

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    Science Advisor
    Homework Helper

    hi tatterspwn! :smile:
    your vi for 90° looks ok,

    so your vix and viy for 35° should also be ok

    but you need to find t for 35° before going any further :wink:
     
  5. Oct 4, 2012 #4
    For that I did
    Tt=2 (Vsinθ)/g
    Tt=2 (6.552sin35)/9.8
    Tt= 0.7670s
     
  6. Oct 4, 2012 #5

    tiny-tim

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    Science Advisor
    Homework Helper

    ah now i see, you did …
    … yes, that seems fine :smile:
     
  7. Oct 4, 2012 #6
    Thank you :approve:
     
  8. Oct 5, 2012 #7
    There remains a problem here. The initial velocity of 6.55 m/s is correct. The time of flight is correct also. However you are mixing up the Vx and Vy velocities. So again I ask how a projectile shot at an angle of 35 degrees from horizontal can have a vertical velocity greater than its horizontal velocity? You have written:

    Vx=3.758m/s
    Vy= 5.367m/s

    Vy cannot be greater than Vx for an angle of 35 degrees. Sin 35 < cos 35.
     
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