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Projectile related question

  1. Jun 12, 2008 #1
    Question 1:
    An aeroplane is flying at a constant horizontal velocity of 50 m s-1 at a height of 1000m. What is its horizontal distance from a target on the ground, so that a parcel released from the plane will hit the target.


    I've tried using the vertical component of motion equation. But the answer is not the same. Help pls.

    Question 2:
    When an aircraft is 4000m above the ground and flying upwards with a velocity of 50 m s-1 at an angle of 30 degree to the horizontal. A bomb is released. Neglecting air resistance, calculate the time taken by the bomb to reach the ground

    I've also tried the vertical component of motion equation. But its wrong too. Help!

    Thanks~
     
  2. jcsd
  3. Jun 12, 2008 #2

    tiny-tim

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    Hi crays ! :smile:

    Show us your working for Question 1, so that we can see where you went wrong, and help! :smile:
     
  4. Jun 12, 2008 #3

    Kurdt

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    If you show what you've done we can better identify where you need help. Essentially what you need to do for this problem is find out how long it takes to hit the ground from the height its released using the vertical component and then find out how far it will travel horizontally in that time.
     
  5. Jun 12, 2008 #4
    Thanks for the reply guys. Here is my working so far for question 1. I've substituted the value into the vertical component. 1000m = 50t + 1/2(10t^2) and finally got a quadratic equation. And i found out t=-20 or t=10. Therefore t=10 is accepted right?

    Then i proceed to applying the horizontal component of motion. x = (u cos Q)t

    i know the initial velocity, u = 50 m s-1
    and the t=10

    but cosQ should be 90 right? Since its dropping vertically down. I'm stucked here.
     
  6. Jun 12, 2008 #5
    no initial velocity in vertical axis is 0...
    the only motion initially is 50m/s in horizontal axis
     
  7. Jun 12, 2008 #6

    Kurdt

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    For the horizontal component don't worry about the angle. Just work out how far it will travel at that speed in that time.
     
  8. Jun 12, 2008 #7

    tiny-tim

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    No, the initial vertical component is not 50, is it? :smie:
     
  9. Jun 12, 2008 #8
    Okay, so i get 500m. But the answer states 714m. Where have i possibly went wrong?
     
  10. Jun 12, 2008 #9

    Kurdt

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    Thats where you went wrong. I missed that :redface:
     
  11. Jun 12, 2008 #10
    I'm sorry. But my physics teacher did not explain it clearly (even after i asked her). I can't really understand the book. Its not CLEARLY stated out. Someone mind to guide me? for the vertical component, i just calculate everything in the vertical direction? but in the question, the only initial velocity given is 50m s-1 no?
     
    Last edited: Jun 12, 2008
  12. Jun 12, 2008 #11

    Kurdt

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    When the plane first drops the parcel, it has no vertical velocity since the plane is traveling at a constant height. Once the parcel is released its speed increases from 0 to its final speed because it is being accelerated by gravity. So initial vertical speed is 0m/s.
     
  13. Jun 12, 2008 #12
    thanks for the reply. Even with that
    1000m = (0)t + 5t^2
    t=14.142s

    then i used
    u(t) = s

    50(14.142) =707.10

    still not the same as the book's answer.
     
  14. Jun 12, 2008 #13

    Kurdt

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    Use 9.81 ms-2 for the acceleration due to gravity.
     
  15. Jun 12, 2008 #14
    Thank you very much. What carelessness i've made.

    what bout the second question?
    Question 2:
    When an aircraft is 4000m above the ground and flying upwards with a velocity of 50 m s-1 at an angle of 30 degree to the horizontal. A bomb is released. Neglecting air resistance, calculate the time taken by the bomb to reach the ground
    (Assume g = 10 m s-2)

    After solving the first question, i tried using the same method on my second question.
    Firstly by using the vertical component
    4000m = (0)t + 1/2(10)t^2
    then i found t=28.28s

    but the answer says the time is 31 seconds.
     
  16. Jun 12, 2008 #15

    Kurdt

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    This time the initial vertical component of velocity is not zero is it since the plane is traveling up at an angle.
     
  17. Jun 12, 2008 #16
    so first i find the vertical component with 50(tan 30) = initial vertical component of velocity ?
     
  18. Jun 12, 2008 #17

    Kurdt

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    Yes almost. I wouldn't use tan though.
     
  19. Jun 12, 2008 #18
    why? wouldn't tangent be easier since you have both adjacent and opposite value but don't have the hypotenuse ? I've tried substituting the value into it.

    4000m = (28.87)t + 5t^2
    t = -31 or 25 (31 is there, but how can it be a negative value?)
     
  20. Jun 12, 2008 #19

    Kurdt

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    The velocity is the hypotenuse. You have to be careful with signs as well. You have defined downwards as your positive, so since the parcel will be initially traveling upwards you need a minus sign in front of its initial velocity.
     
    Last edited: Jun 12, 2008
  21. Jun 12, 2008 #20
    Oh i totally forgot that velocity is going diagonally according to the angle, i've been thinking its traveling in the horizontal way. Thanks too for the reminder of signs, i've been careless.

    4000m = -25t + 5t²
    t = 31 or t = -25. Thanks!
     
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