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Homework Help: Hight reached by a projectile

  1. Apr 9, 2010 #1

    DR1

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    firstly i am new here so be nice please

    the problem i have is that i am stuck on a question in my assignment which is

    a bullet of mass 0.6kg is fired vertically upwards by a vertical propelling force of 50N over a time scale of 0.5 seconds . calculate

    1)the velocity of the bullet ehen it leaves the gun. ( the answer i have for this is based on a=f^m ie 50^0.6 giving a= 83.3 ms-2 then using V=U+AT therefore v=0+83.3X0.5 getting the final answer of V= 41.65ms-1) can you just confirm if i am on the right lines with this first

    2) The Height reached by the bullet ( again i have attempted this by reaaranging v^(2)=u^(2) +2as for s to get 10.4125m)

    3) The time taken from the bullet leaving the gun to its return to the same height ( i am stumped at to were to start with this one i know that gravity is going to have its effect on the return and that that will be related to the above height answer)

    My main issue is that dispite getting what look to be answers to my questions i keep doubting/questioning myself

    any help would be appreciated i am not looking for you to give me the answers just some guidence would be more than helpfull

    many Thanks
     
    Last edited: Apr 9, 2010
  2. jcsd
  3. Apr 9, 2010 #2
    Hey I just checked out number 1 and you are on the right lines but you are forgeting a force in your calculation. There is one more force that you need to account for. Let me look at 2 and 3
     
  4. Apr 9, 2010 #3
    Ok number 2. I'm not sure how you did that. What equation are you using and what is s? s is time?

    Use the kinematic equations [tex]y_{0}=y_{i}+v_{i}t+\frac{1}{2}at^{2}[/tex]
    and v[tex]_{0}=v_{i}+at[/tex]

    Using the velocity equation know the velocity will 0 when it reaches its max height right? it will stop, and we calculated vi in number 1, and after the bullet leaves the gun, there is no more 50N propulsion so its just gravity acting on it. So a is 9.81m/s^2.. So we can find time.
     
  5. Apr 9, 2010 #4

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    for question 1 is the other force you are refering to gravity ie 9.81

    and in question 2 s is distance if that makes any difference

    in the equation you have given for question 2 there is a y what does this correspond to
     
  6. Apr 9, 2010 #5
    oh ok, then sorry for my ignorance, whats u? I thought u was distance but now that I look at it, that wouldn't make sense. Perhaps u is time?

    Anyways, if it works when you change your accel, you should get a height of 68.87m.
    That equation looks familiar, I have just never used it. But it doesn't matter which one you use.

    As for number three, you can/should calculate it, but is something that can be known intuitively if you have learned conservation of energy.
    To calculate it, using kinematics, you know your vi is 0 at the top, you know your height, you know the accel, so solve for t and you got it.
     
  7. Apr 9, 2010 #6

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    v=final velocity (41.65ms-1)
    u=initial velocity (o ms-1)
    s=distance
    t=time
    a- acceleration (83.33ms-2)

    i rearanged the formula to get
    S=V^(2)-U^(2)
    ------------
    2a
    this gets me the answer of 10.41m

    i am obviously missing something if 68.87m is the answer
     
  8. Apr 9, 2010 #7

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    has anyone else got any ideas as im still a little perplexed
     
  9. Apr 9, 2010 #8
    hi,
    If you break the problem up into a first stage where the bullet is fired from the gun to get the exit velocity, then treat finding the height as a seperate problem.

    As frozenguy says at the maximum height your final velocity will be zero (v=0) and the acceleration will be gravity (a=-9.81), the initial velocity will be the result fo the first part of the question (u=v in the first part)

    Hope that helps
     
  10. Apr 9, 2010 #9

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  11. Apr 9, 2010 #10
    that is the correct equation for the change in height with constant acceleration,

    although i havent checked that your value for the answer to part a is correct i.e. 41.65.
     
  12. Apr 9, 2010 #11
    This is your problem. It started in problem one. You forgot to add one force. And you even were correct about it in this post. (although, 9.81 is not a force, its the magnitude of the acceleration due to gravity. The force of gravity is the mass multiplied by the accel due to gravity.)

    Go back to your first problem and recalculate the acceleration taking into considering the force of gravity on the bullet, not just the force of propulsion. Gravity is acting on the bullet at all times.

    Your acceleration should be ~ 73.5[m/s^2]
     
  13. Apr 11, 2010 #12

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    right so now my answers are as follows

    acceleration is 73.5 due to gravitys effect on the acceleration i get that
    so my velocity when leaving the gun as per question 1 would be 36.75ms-1 using V=U+AT

    therefore using S=U^(2)-V^(2) DIVIDED BY 2g
    i get the answer of 68.84m as height reached by projectile (question 2)

    now for question 3 i am tying myself up a bit i have used 2 different formulae but not sure either is correct as i get different answers the first i used was t=(v-u)/a giving me an answer of 0.5s for up but does this double to give up and down time?? alternatively i tried the t=s/v which gives 1.87s for up. which is right? and is the bullet traveling as fast back down again

    i appriciate you have all tried to explain this in previous post but im still confused im afraid and can someone comfirm my answers for 1+2 are looking correct please

    many thanks

    Dave
     
  14. Apr 12, 2010 #13

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    anybody got anything to add
     
  15. Apr 12, 2010 #14
    I'm just not familiar with the formulas you are using, have you used the regular two kinematic equations? Those are very easy to use.

    The time it takes to reach its max height is the same time it takes to fall back down.

    So double the time it takes to fall back down.

    you can also use the kinematics

    Y0=yi+vit+.5at2

    y0= 0 because 0 is where you started.
    yi=max height because youre analyzing from max height to 0.

    vi=0 because the bullet stops moving at ymax

    So you are left with yi=.5at2

    You know a, you know yi, then solve for t.

    And yes, I got 68.87 which is only a .04% error so you should be good to go on 1 and 2 unless I messed up :P
     
  16. May 18, 2010 #15
    Dave / DR 1

    Please contact me on bomber284@live.com. I am just finishing off TMA 2 now and we may be able to help each other in the future. Looks like you had a few probs with the Mechanical stuff, dont we all, i wouldn't worry about it. The reason we are doing Electrical Engineering is because we are Electrical biased not Mechanical. What TMA are you on now, plus i think this is a US site, so maybe the help was coming from US guys not Brits, hence the reason the equations are different.........only a guess. Look forward to hearing from you.
     
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