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Projectile Motion velocity homework

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    A projectile is fired horizontally from a gun that is 55.0 m above flat ground, emerging from the gun with a speed of 350 m/s.
    a) How long does the projectile remain in the air?
    b) At what horizontal distance from the firing point does it strike the ground?
    c) What is the y-component of the velocity as it strikes the ground?
    d) What is the magnitude of its speed as it strikes the ground?
    e) At what angle does the projectile strike the ground?

    2. Relevant equations
    V1x=V1cosθ
    V2x=V1cosθ
    x=(V1cosθ)t
    V1y=V1sinθ
    V2y=(V1ysinθ)t-gt
    y=(V1sinθ)t-0.5gt^2


    3. The attempt at a solution
    So I attempted each one and just want feed back on my answers.
    a)t=3.4s
    b)x=1190m
    c)-33.32m/s
    d)351.6m/s
    e)355 degrees
     
  2. jcsd
  3. Feb 3, 2013 #2
    Your method is correct, but.

    1. The precision of the time result is rather low, so some of the significant digits in the subsequent results are incorrect.

    2. 355 degrees cannot be an angle with which the projectile strikes the ground, because this is underground. The angle must be less than 90 degrees.
     
  4. Feb 3, 2013 #3
    So the timing is incorrect?
    This is what I did....
    y=(350m/sin(0))t-0.5gt^2
    -55m=-0.5(9.8m/s^2)t^2
    -55m/-4.9m/s^2=t^2
    t=3.35s so I rounded it to 3.4s
     
    Last edited: Feb 3, 2013
  5. Feb 3, 2013 #4
    Not necessarily.
     

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    Last edited: Feb 3, 2013
  6. Feb 3, 2013 #5
    So instead of adding 360 degrees, I should have added 180 degrees?
     
  7. Feb 3, 2013 #6
    Im not sure of you question. Adding 360 to 355 would get you to 355 again, you would make a complete circle. If I have interpreted your answer correctly, 355 into the ground, equals 175 above y axis. I have not worked the problem though.
     
  8. Feb 3, 2013 #7
    3.35 s is better.

    For an angle with a surface, one may may always choose either an angle greater than 90, or less than 90. Usually the smaller angle is selected.
     
  9. Feb 3, 2013 #8
    Okay so I took the tanarc(-33.32/350) and got -5 degrees... Then I added 360 to get 355 degrees.... So I would add 180 degrees instead of 355 because 355 degrees would be the angle it penetrates the ground? and 175 degrees would be the angle at which the projectile hit the ground?
     
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