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Projectile and Horizontal Distance

  1. Jan 25, 2008 #1
    [SOLVED] Projectile and Horizontal Distance

    1. The problem statement, all variables and given/known data

    A projectile launched at an angle [tex]\theta[/tex] to the horizontal reaches a maximum height h. Show that its horizontal range is 4h/tan[tex]\theta[/tex]

    2. Relevant equations

    Can you help me figure out how to get this setup properly and what equations i should use?

    3. The attempt at a solution

    I can not figure out how to answer this question i have tried deriving my own forumla and none of them are coming out correctly
    Last edited: Jan 25, 2008
  2. jcsd
  3. Jan 25, 2008 #2

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    Find the vert and horz components of the initial velo. Then find max height reached using vert comp as the initial velo upward. That'll give you a relation between h and theta. Now find the total time of flight until vert dist is zero again, and from that the horz dist traveled in that time.

    If you don't go through your books or notes, and write what you have done next time, it'll be difficult for us to help you.
  4. Jan 25, 2008 #3
    i have been going through my notes and online hw problems and looking online and reading the book and looking at his lecture slide shows, which blow... he dosnt teach we have to teach ourselves and work in groups.. my group couldnt figure it out either,

    i dont get how i am soposed to get the vertical and horiz component

    i got for the first part of this solution

    tan theta = h / q if q if the horizontal axis

    i have equation i could use but dont really see how to apply them

    vector v = vector v not + vector a*t
    vector r = vector r not + vector v not*t + 1/2 vector a*t^2

    v sub x = v sub x not
    v sub y = v sub y not - g * t
    x = x not + v sub x not * t
    y= y not + v sub y not * t - 1/2 * g * t^2

    projectile trajectory
    y = x tan theta - g / ( 2 *v not ^2 * cos ^2 theta not ) * x^2

    horizontal range
    x = v not ^2 * sin 2 theta not / g

    theta = tan^-1 A sub y / A sub x
    A sub x = A cos theta
    A sub y = A sin theta

    my last problem left due for next week and it's killing me! lol
  5. Jan 25, 2008 #4

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    (Post above deleted by OP.)

    (You need not write v not, but write v0. The not is not "not" here, but nought. :smile:)

    The problem with you is that that you know nought about projectile motion. It's not possible for us to write down the whole chapter explaining the concepts, or even write each of the eqns. Can you do some more studies with solved examples on your own?

    Let's see what you can do in a couple of days, and then come back here. Next week is far away. :cool:
    Last edited: Jan 25, 2008
  6. Jan 25, 2008 #5
    next week as in Monday..
  7. Jan 25, 2008 #6

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    Do you know the eqns of st line motion under uniform accn, like vertical motion under gravity?

    Do you know how far a particle goes in time t if projected witha velo v in absence of gravity?
  8. Jan 25, 2008 #7
    yes.... of course...
  9. Jan 25, 2008 #8

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    How much would a body rise up vertically if thrown up with a speed of 'u'? How much time would it take? What's the speed up at that highest point?

    What's the answer to my 2nd Q? I wanted you to answer with the answer.

    (There was no reason to delete your original post after post #2! At least that showed that you've gone through some mathematics.)
    Last edited: Jan 26, 2008
  10. Jan 27, 2008 #9
    well if your just gonna assume im not trying to solve this problem and make fun of some information i gathered considering im only a week or so into physics so i dont have equations and equation names memorized yet... i have talked to half a dozen people that havnt figured this problem out..

    these are the equations i believe i can use to solve the problem...

    accel is vertical and const downward
    x=x0 + v0x t
    vx = v0x

    vertical motion has grav. accel. vertical motion is free fall
    y = y0 + v0y t - 1/2 g t^2
    vy = v0y - g t

    trigonometric relations between components
    v0x = v0 costheta
    v0y= v0 sin theta

    ymax = (v0^2 * sintheta^2) / (2 * g)

    ugh this is insane trying to achieve this relation
    Last edited: Jan 27, 2008
  11. Jan 27, 2008 #10

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    On the contrary, you are going to need answers to exactly these questions to solve this problem. So you may understand that I had no intention of making fun of you. BTW, who were those half a dozen people?

    (I’m writing in shorthand.)

    The horz comp of the velo remains const = v*cos(theta). The vert comp of the velo undergoes the change.

    Initial vert comp of velo = v*sin(theta)

    If h is the max height, then 0 = (vsin(theta))^2 – 2gh (1)

    If it takes time T to reach max height, then,
    0 = vsin(theta) – gT => T = vsin(theta)/g. –(2)

    The total time taken to fall back to ground again is 2T.

    Horz dist covered in time 2T
    = vcos(theta)*2T
    = vcos(theta)*2vsin(theta)/g [from (2)]
    = 4h/tan(theta) [using value of v^2 from (1)].
  12. Jan 27, 2008 #11
    i actualy had equation one, and i couldnt find how to relate time... i talk to my physics group and part of another group, and they said they got something and it didnt make sense or they were lost

    i understand everything you did completely and it makes perfect sense, thanks for the help

    i just hate people that ask me for help when they dont do ****, so i never ask for help until i have tried everything and put in a lot of time, i had a calc friend who would always come to me complaining he couldnt figure something out, turns out he put like 10 mins into it and gave up, i have prolly put about 10+ hours into this problem lol ugh.. i dont like to rely on people, even though i post on here im not waiting for a response im still working looking through all my crud on my desk

    i know you gotta be the same way you probely get a lot of plp that post w/o putting a lot of effort into solving it on there own, i kinda take it personally when someone tells me to do more when i know i have done everything that is available to me
  13. Jan 27, 2008 #12
    He's not breaking your balls. Look at your initial post. Does it look like it was posted by someone who has worked on it for 10+ hrs? Maybe you should spend 9 hrs doing the problem and another hour posting everything you've done. Don't get too revved up, you set yourself up.
  14. Jan 27, 2008 #13
    Yeah do that next time! A lot of people here are obviously smart and by just analyzing your work, they can spot the mistake w/o having to go thru the calculations themselves.
  15. Jan 29, 2008 #14

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    A number of intermediate posts deleted by the OP. The OP should be warned against doing this.
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