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A good projectile question

  1. Jun 22, 2013 #1
    Hi,

    How does one prove that a projectile will shoot 3x higher at 60 degrees that at 30 degrees?

    And there is no more info available.

    Here is what i attempted:

    I gave it 500m/s,

    1a.) The formula for maximum time @ 60 degrees: th = (u)(Sin60)/9.8
    = (500)(Sin60)/9.8
    = 44.185's

    2a.) The formula for maximum height @ 60 degrees: h = (u)(t) + (1/2)(a)[(t)"squared"]
    = (500)(44.185) + (1/2)(9.8)[(44.185)"squared"]
    = 31658.8397m (31.659km)

    1b.) The formula for maximum time @ 30 degrees: th = (u)(Sin30)/9.8
    = (500)(Sin30)/9.8
    = 25.510's

    2b.) The formula for maximum height @ 30 degrees: h = (u)(t) + (1/2)(a)[(t)"squared"]
    = (500)(25.510) + (1/2)(9.8)[(25.510)"squared"]
    = 15943m (15.944km)

    3a.) Now if i devide height of 60 degrees with the height of 30 degrees
    I get 31658.8397m / 15943m = 1.986 times the height and not 3x the height

    Did i do it right or am i looking at it in a wrong way?

    Regards
     
  2. jcsd
  3. Jun 22, 2013 #2
    Your method is fine. It must be the calculations. I would suggest doing it with variables instead of specific values - that way you don't make errors calculating.

    EDIT : I noticed you are using value of u as 500m/s in the distance equations. That is incorrect - you must use the upward value of velocity, which is u*sin*theta (theta = initial angle).
     
  4. Jun 22, 2013 #3

    tiny-tim

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    Welcome to PF!

    Hi HadanIdea! Welcome to PF! :smile:

    Nobody asked you for t, so why are you bothering to find it??

    Use another constant acceleration equation, that doesn't involve t ! :wink:
     
  5. Jun 22, 2013 #4

    SteamKing

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    In equations 2a. and 2b. in the OP, you use the initial velocity of 500 m/s in calculating the max. height of the projectile. This is incorrect. You must use only the vertical component of the initial velocity in calculating the max. height of the projectile.
     
  6. Jun 22, 2013 #5
    Don't put in sample numbers at an early stage; look at the big picture first and find proportionalities.

    Let "_v" denote vertical, "_h" denote horizontal components, "∝" means "is proportional to"

    Decompose momentum
    p =: p_v + p_h,

    consider Energies
    V ~ mgh => V ∝ h
    T = p²/2m

    and think about
    - if g affects p_h
    - how to decompose T into T_v and T_h
    - and think about p_v and T_v at the top of the trajectory.
     
    Last edited: Jun 22, 2013
  7. Jun 22, 2013 #6

    TSny

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    HadanIdea,

    My 2 cents advice is to heed tiny-tim's 2 dollar advice!
     
  8. Jun 23, 2013 #7

    PeterO

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    When resolving the 30 degree - 60 degree launch angles, you will be resolving the launch speed using the familiar 2-1-√3 triangle.

    This shows that , compared to a 30 degree launch, the vertical speed is larger by a factor of √3.

    That means the average vertical velocity before reaching maximum height is higher by a factor of √3.
    The time taken to slow to 0 vertical velocity is also up by a factor of √3

    So you travel at √3 x the speed for √3 x the time.

    What result will that give?
     
  9. Jun 23, 2013 #8

    TSny

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    Nice. :cool:
     
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