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Projectile motion

  1. Jun 24, 2011 #1
    1. a rescue plane is flying at a constant elevation of 1.2km with a speed of 430km/h toward a point directly over a person struggling in the water. at what angle of sight should the pilot release a rescue capsule if it is to strike the person in the water...answer is 57.3degreen from vectical.



    2. i dont know what the question talking about. can someone explain to me( with diagram is better)?? thank you



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 24, 2011 #2
    I'd make you a diagram if I had an art program on this computer, but alas . . . here are your variables. Always remember to write your variables. It makes visualization so much easier (for me, at least).

    [itex]h = 1.2km = 1200m[/itex]

    [itex]v = 430km/h = 430000m/3600s \approx 120m/s[/itex]

    And the question is basically asking:



    Keep in mind that the crate is initially moving at the same velocity as the plane, so it will carry forward when it's released. That's why there needs to be an angle of difference between the mark and the plane.

    Let me know if you're with me still.
     
    Last edited: Jun 25, 2011
  4. Jun 25, 2011 #3
    what is the speed i should take?
     
  5. Jun 25, 2011 #4
    Use the speed 120m/s, but just know that I rounded that speed up from 119.444m/s.

    Remember this FOREVER:

    [itex]KMS[/itex]

    [itex]Kilogram, Meter, Second.[/itex]


    If this helps you remember:

    [itex]Kiss, My, Sock[/itex]


    Always use these units.
     
  6. Jun 25, 2011 #5
    no~i mean i should take vertical speed or horizontal speed?
     
  7. Jun 25, 2011 #6
    You need to investigate both.

    Imagine dropping a ball while standing still and think about how it moves. Compare that to the motion of a ball dropped while running.

    If you drop a ball while standing still, the only noticeable force acting on it is the gravitational acceleration.

    And, if you drop a ball while running, the ball's path is again effected by gravity (like before), but now also your horizontal velocity.

    In the second example here, both horizontal and vertical velocities change the position of the ball. Just like with the crate dropped from the plane.
     
  8. Jun 25, 2011 #7
    Can I see your work on this question so far?
     
  9. Jun 25, 2011 #8
    okay
    1st i find vertical speed. vertical speed = 119.44/tan angle
    2nd i find angle, i use this formula, v^2= u^2+2as, 0= (119.44/tan angle)^2 +2(9.8)(1200)
    angle= 37.91degree from vertical
    and the answer is wrong~
     
  10. Jun 25, 2011 #9

    rl.bhat

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    Homework Helper

    Find the time t taken by the parcel to cover 1200 m height to reach the ground. Take initial vertical velocity of the parcel zero.
    Find the horizontal distance covered by the parcel before it hits the person, using x = horizontal velocity x time.
    Then tanθ = x/h
     
  11. Jun 26, 2011 #10
    how to find the vertical velocity?
     
  12. Jun 26, 2011 #11
    thank you, i got it.....
    thank you very much......
     
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