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Homework Help: Projectiles Question

  1. Jun 26, 2007 #1
    When a aircraft is 4000m above the ground and flying upwards with a velocity of 50 m s-1 at an angle of 30° to the horizontal. A bomb is released. Neglecting air resistance, calculate
    a) the time taken by the bomb to reach the ground,
    b) the velocity of the bomb when it strikes the ground. (Assume g = 10 m s-2)

    The answer given by the book is:
    a) 31.0 s
    b) 264 m s-1, 9.5° from the vertical downwards

    How do you get those answers? Please help me.

    I have tried to get those answers but I only managed to get the first one. I used s=ut+1/2at^2 for the vertical component to get the time. Since u=0 for vertical component, -4000=-1/2(10)t^2 will get t=30.9s.

    Then for the second question, I got the horizontal velocity and vertical velocity when t=30.9s and using a vector diagram and the pythagoras theorem, I got -287 ms-1 for the velocity. However, the answer that was given by the book says it should be 264 m s-1. Is my answer correct or the book's answer correct?
    Last edited: Jun 26, 2007
  2. jcsd
  3. Jun 26, 2007 #2
    heh...I got stuck at where you are also; the vf I received was higher; are you sure the 6 in 264 looks nothing like an 8...then it'd be reasonable (I'm just an amateaur so I cant help too much).
  4. Jun 26, 2007 #3
    I am sure that the answer in the book is 264 ms-1 and 95 degree. My answer deviated from both of them.
    Last edited: Jun 26, 2007
  5. Jun 26, 2007 #4
    My answer comes out as vf = ~283.945m/s, in which vf could be estimated to 284m/s, which is somewhat like 264m/s, just with the 8 instead of the 6. The formula I'm using is vf^2 = vi^2 + 2a(deltay). Sorry, I'm not much of any help, and the only conclusion I have is that the book may be wrong (although that usually isn't the case).
  6. Jun 26, 2007 #5


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    u is not zero for part a)

    The bomb is launched at an angle of thirty degrees w.r.t. the horizontal at a speed of 50 m/s.

    This means that the bomb has a vertical component upwards, which is u.
  7. Jun 26, 2007 #6
    Oh sorry for my initial post stating that I used u=0. That was for a different question. I got my workings messed up as I hurriedly tried to find them after I found out that you need to post your workings.

    I used u = 50sin30 = 25ms-1 and subbed it into s=ut+1/2at^2 to get t.
  8. Jun 26, 2007 #7
    Draw a picture to simplify the problem.

    Vertically, you have;

    s = 4000m \hfill \\
    u = 50\sin 30\;ms^{ - 1} \hfill \\
    a = - 9.8\;ms^{ - 2} \hfill \\
    \hfill \\
    s = ut + \frac{{at^2 }}
    {2} \hfill \\
    4000 = 50\sin 30t - 4.9t^2 \hfill \\
    0 = - 4.9t^2 + 50\sin 30t - 4000 \hfill \\
    0 = (t + 26.13)(t - 31.236) \hfill \\
    t{\text{ must be positive:}} \hfill \\
    t = 31.236\;\sec \hfill \\
  9. Jun 26, 2007 #8
    another note as mentioned:
    u = 50\sin 30\;ms^{ - 1} = 50\left( {\frac{1}
    {2}} \right) = 25ms^{ - 1} \hfill \\
    {\text{If you do trig or calc; exact values, or just plug into calculator}} \hfill \\
  10. Jun 26, 2007 #9


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    s should be -4000 meters (use g =10).
  11. Jun 26, 2007 #10
    how does that work??

    the magnitude of the vertical displacement is positive from the ground
    the acceleration due to gravity is pulling down on the plane, i.e. in a negative direction
  12. Jun 26, 2007 #11
    Yeah but the bomb goes down and causes the displacement to be negative since it is 4000m below the initial point of drop. I really need to know if the bok's answer is correct or is it wrong. My physics teacher got 287ms-1, one of my classmates got 264ms-1 using an unorthodox method of vector diagram which he cannot explain why he used so and so values but he got the exact value in the book. He said he used the cosine rule.
  13. Jun 26, 2007 #12
    What is the book you're using?
    Last edited: Jun 26, 2007
  14. Jun 27, 2007 #13


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    When time is zero the bomb is released from the aircraft. If you take the ground as the reference point the equation will be

    [tex]s = 4000 + ut - \frac{1}{2}gt^2[/tex]

    in this case one will solve for s = 0 when it hits ground or

    [tex]0 = 4000 + ut - \frac{1}{2}gt^2[/tex]

    if one takes the release point as the reference the equation will be

    [tex]s = ut - \frac{1}{2}gt^2[/tex]

    since up is chosen as positive the displacement will be -4000 when it hits ground giving

    [tex]-4000 = ut - \frac{1}{2}gt^2[/tex]

    which again give the same equation to be solved as with the other reference point (ground)
  15. Jun 27, 2007 #14
    yeh true 'andrevdh'
    suppose that method is more 'physically correct'
    but they give the same answer none-the-less

    thanks for pointing that out ...
  16. Jun 27, 2007 #15
    So? What's the answer?
  17. Jun 29, 2007 #16
    the world may never know; although...it might be possible its marginal error (a very big) considering your friend got the answer using another method. I'm still getting another answer.
  18. Jun 29, 2007 #17
    What about the experts? I joined this forum hoping that some of the more advanced physics students could enlighten me in a situation like this. Well, for those who posted here, thanks anyway. At least I know someone somewhere in the world got the same answer as me.
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