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Air resistance physics

  1. Nov 12, 2006 #1
    Hi guys, got a question and been trying to solve it for 2 days.

    the question is:

    In the absence of air resistance with what angle to a horizontal ground must an object be thrown from ground level so that it ravels a distance before landing equal to twice the maximum height ?

    I know that i've got to work with simaltaneius equations, tried all the linear equations but can't get to an answer. Any one can show me the steps to arrive to the answer pls ?

    Thank you in advance
     
  2. jcsd
  3. Nov 12, 2006 #2

    rsk

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    Set up an equation to calculate the maximum vertical height. Use vf^2 = v0^2 + 2 a s

    At this point the projectile has travelled how far horizontally?

    Can you equate these now to solve for the angle?


    Edited......haven't tried this yet but you might need x = v0t + 0.5 a t^2 for the vertical instead, to be able to solve...
     
    Last edited: Nov 12, 2006
  4. Nov 12, 2006 #3
    This might help: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra14

    [tex]\frac{h}{R} = \frac{\tan \theta}{4} = 1/2 [/tex]​

    Then, you may find theta.

    Note: You should not memorize this formula, but understand how to deduce it.
     
  5. Nov 12, 2006 #4

    i tried that mate still didn't manage to. i'll show you what i did.

    Vertical:
    s=ut+0.5at^2
    s= usinΘ*t + 0.5(-10)(t^2)
    s=usinΘt+5t^2

    Horz:
    s=ut
    2s=ucosΘ*2t (2t because maximum height =t, and it takes double that to go down again, 2s because double the distance)

    then

    tried to substitute but can't come to a conclusin.

    I also tried s=ut+0.5at^2 but still can't :S
     
  6. Nov 12, 2006 #5
    I'll try to use that. Thnaks to that i know that the answer should be 63.43494882292201 :rolleyes:
     
  7. Nov 12, 2006 #6

    rsk

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    Yes, sorry, I'm still trying too and can't get it.

    Each time I try I either end up with lots of unknowns, or else everything cancels out to leave something trivial.

    WIll post back when I get somewhere!
     
  8. Nov 12, 2006 #7
    thx mate, i've spent at least 4 hours trying to work them out, but always got a lot of unknowns. Mainly the time keeps me away from solving it,

    So i thought i would get out an equation for time, but then i got like a cos^-1 of 0 which is not possible.
     
  9. Nov 12, 2006 #8

    rsk

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    got it

    In the vertical, use v^2 = u^2 + 2as to find an expression for the max height.

    I got h = (usina)^2/19.6 ------------------(1)

    (a of course is the angle, I don't have alpha or theta!)

    Then use x = ut + 0.5at^2 to find an expression for T, the time for the object to land (ie vertial displacement = 0).

    I got T = usina/4.9

    Then in the horizontal, the obhect travels 2h in this time T

    so 2h= ucosaT

    or h = (ucosa)T/2 ---------------(2)

    I then equated these 2 expression for h, and after a bit of rearranging got

    sina/cosa = 19.6/9.8

    or tana = 2



    Try it and check there are no stupid errors......
     
    Last edited: Nov 12, 2006
  10. Nov 12, 2006 #9
    You have that the equation of motion in 2 dimensions is

    [tex] \mathbf{r}(t) = (v_0\cos(\theta)t) \mathbf {\mathrm{i}} + (-gt^2/2 + v_0\sin(\theta)t)\mathbf{\mathrm{j}} [/tex]

    Where [itex] \theta [/itex] is your firing angel to horizontal ground. From here you can find an expression for how long the projectile travels horizontally. Because what is the vertical component of the vector [itex] \mathbf{r} [/itex] equals to, when the projectile hits the ground? When you know that, you can solve the vertical equation for the time. You can then substitute that expression for the time in the horizontal component of the vector to find out how far along the horizontal the projectile traveled.

    To the find the maximum heigh above the ground level, you have to use the assumption that the vertical component of you velocity vector [itex] \mathbf{\dot{r}}(t) [/itex] is equal to zero at the maximum height. Solve that equation for the time, and when you substitute that expression for the time into your vertical component of the vector [itex] \mathbf{r}(t) [/itex] you find the height, and for this time it will be te maximum height.
     
  11. Nov 12, 2006 #10

    rsk

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    Given that what I ended up with is basically tana = 2g/g, i wouldn't be surprised if there was a quicker way to do it.....
     
  12. Nov 12, 2006 #11
    Understand it. But can get how you got

    "Then use x = ut + 0.5at^2 to find an expression for T, the time for the object to land (ie vertial displacement = 0).

    I got T = usina/4.9"

    could you please give me the steps to arrive to that usina/4.9
     
  13. Nov 12, 2006 #12

    rsk

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    at this time the vertical diplacement is =0 - it's back at the level it started.

    so 0 = ut + 0.5 a t^2

    so 0 = usina T - 4.9 T^2

    so usina T = 4.9 T^2

    cancel T from each side

    u sina = 4.9T

    so T = u sin a / 4.9
     
  14. Nov 12, 2006 #13
    thanks rsk for explaining everything :)

    worked everything, and it's 100% correct.
     
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