1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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!
     
  22. Jun 12, 2008 #21
    sorry Kurdt i have another question here, related to the previous question too.

    it asked for the velocity of the bomb when it strikes the ground. i used the formula v = u + at. Since i have u (-25) , a (g=10m s-2) and t = 31. Is the the right thing i should do?
     
  23. Jun 12, 2008 #22

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, but remember that the bomb will have a horizontal component of velocity aswell.
     
  24. Jun 12, 2008 #23
    okay... what am i suppose to do with the horizontal component of velocity ?
    I thought of finding the displacement with R = (u² sin 2Q)/g which is

    R = (50² sin60)/10
    =216.5m

    then use
    216.5 = 50t + 5t²
    find the t and use
    v= u + at

    then merge the v = u + at of the vertical with the horizontal?
     
  25. Jun 12, 2008 #24

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Horizontal is easy, its just the plane's component of velocity in the horizontal direction. Since, there is no air resistance that will be the answer.
     
  26. Jun 12, 2008 #25
    sorry but could u explain that abit? so i just use the formula R = (50² sin60)/10 to find the distance ? That's all ?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook