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Aeroplane Projectile

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/known data

    A windscreen is dirty and a bird dropped something which hits perpendicular your windscreen (Which has an inclination angel 40°). By comparison with the height of your house you know the flight height of the bird quite well. 10 m above your car. Determine the Horizontal flight velocity of the bird? we are ignoring friction.... Find out also.....Time for falling 15m... Vertical velocity after dropping 15m.......Required Horizontal velocity



    2. Relevant equations and attempt
    Time for free fall after dropping 15m

    y= Vot - 0.5 x 9.81 t^2 (g= -9.81m/s2) (y=15m) (Vo= initial velocity is 0)
    t= 3s. ( by calculation)

    Vertical velocity after dropping 15m.........

    v= Vo + gt
    = 0 + 9.81 m/s^2 X 3s
    = 29.43s





    3. The attempt at a solution
    but I think my method is wrong.

    I can use also by this way,
    Horizontal velocity Vx = Vxo
    Vertical velocity Vy = Vyo - gt

    Basically i am not sure that my method to solve the question is ok
     
  2. jcsd
  3. Feb 16, 2015 #2

    BvU

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    Hello Ahmad, welcome to PF.

    This is a nicely different exercise from the usual projectile problems. You really want to sort out things and make a plan how to come to the solution. A drawing would be very helpful.

    What do you know ?

    From the exercise text:
    The dropping is from a height of 10 m. Not 15 m as you seem to think (why?)
    The angle of the trajectory at the moment of impact is 50 degrees wrt horizontal ( ##\perp## the 40##^\circ## windscreen ).​

    From your textbook or your lecture notes, or from here :
    In the horizontal direction there is uniform motion with constant speed( horizontal v0 of the bird ).
    In the vertical direction there is uniformly accelerated motion with constant acceleration ##\ \vec g = -9.81\ ## m/s, starting with 0 m/s as you stated.​

    Combine all these goodies in a few compact equations. Your start (find t) is the right approach. And you do already have the right expressions (partly in numbers, partly in symbol) for the velocities at impact in your attempt at solution. Those give you an angle and bingo ! Good luck.
     
  4. Feb 17, 2015 #3
    you are right but I need the exact solution..............I draw the trajectory and I understand quite well but don't mind. I need the exact solution of Horizontal and vertical velocity.......
    I can compare my solution.............Thanks a lot
     
  5. Feb 17, 2015 #4
    If I consider 10 m height and I get t = 2.03s and vertical velocity = 19.91 m/s. and Horizontal velocity = 0
     
  6. Feb 17, 2015 #5

    BvU

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    Can't be right. The dropping has to be 'launched' with a nonzero velocity and it has to land at a 50##^\circ## angle, not vertically.
     
  7. Feb 17, 2015 #6
    Can you explain please and write the solution. Yes I get what you mean but I really do not know how to get Horizontal velocity...
     
  8. Feb 17, 2015 #7

    BvU

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    And the 2.038 isn't right either. Can you show how you calculate it ? What equation ? What are the dimensions ?
     
  9. Feb 17, 2015 #8
    X= Vot - 0.5 gt^2
    10m = 0 - 0.5 x 9.81 x t^2
    t = 1.43 s.

    Vertical velocity
    V= Vo + at
    =0 - 9.81 x 1.43s
    = -14m/s

    and Horizontal velocity ????​
     
  10. Feb 17, 2015 #9

    lightgrav

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    ok, t for 10m drop is okay, so is the vertical velocity ... 9.81 has units m/s2 , here as an acceleration.
    is the horizontal velocity faster than 14m/s , or slower than 14m/s ? (how does it hit?)
     
  11. Feb 17, 2015 #10

    BvU

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    You now have one component of the velocity at impact on the windshield. The angle at landing is prescribed and that determines the other component !
     
  12. Feb 19, 2015 #11
    I get Vy= 14 m/s and Vx = 11.76 for example, tan 40° = Vy/Vx Vx= 11.76 is it right? last but not least, I am really thankful who help me to solve it out
     
  13. Feb 19, 2015 #12

    gneill

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    The magnitudes look good.
     
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