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A few questions - mostly Kinetic Energy stuff, help please?

  1. Dec 8, 2005 #1
    Hi guys,

    I'm new to the forums so I hope I am putting these questions in the appropriate place. I tried to look around a bit before hand and make sure I found the proper place, but my apologies ahead of time if I am somehow off in that regard. On to the questions.....and thanks ahead of time for ANY help!

    Q#1- You fire 3 shots from your gun (let's label the shots A,B,C). One straight up (A), one straight down (B) and one horizontally (C). Assuming you shoot from the hip (i.e. your hand is in the same place each time, right by your holster) and you ignore air resistance,

    A.) Which of the 3 bullets will have the greatest speed as it reaches the ground and why?

    B.) Now include air resistance, how does your answer change and why?

    **These were questions on a quiz I have already taken and did poorly on, I'm trying to go back over all my quizzes to prepare for next weeks final exam, I also have the opportunity to re-do one quiz and turn it in today!**

    My original answer for part A was that bullet A would have the greatest speed because it's falling a greater distance and can gain speed over that distance. I was using the equation v^2=2gh to show this saying that the "h-height" value would be greatest for bullet A so it had to have the greatest speed. Now, I'm thinking perhaps they all have the same speed? or maybe it's bullet B that is the fastest since it's shot straight down and has the force of the gun directly behind it, but I'm not sure if I am headed in the right direction or just further confusing myself, and I don't know how I can explain either in "physics" terms. Please help if you can offer any advice, suggestions, etc. Thanks!:smile:
  2. jcsd
  3. Dec 8, 2005 #2
    In all cases velocities will be the same
    suppose when it is fired straight down
    final velocity will be v= root ( u^2 + 2gh)
    In the case when it is fired up.....when it will be back to the same point the velocity will be the same as in downward case ( energy conservation)
    so it will aslo take the same time as previos
    third case
    at the time of collision ...horizontal velocity will be unaffected ( since no acceleration of retardation)....and vertical comononet will be root(2gh)....so you can calculate the net velocity...root ( u^2 + 2gh)
  4. Dec 8, 2005 #3
    Awesome, so I was kinda headed the right direction in assuming they would all be the same velocities, but I'm still unsure of how to explain that in physics terminology....

    What is "u" in the equations you use?

    Thanks for your help, it's really appreciated!
  5. Dec 8, 2005 #4
    u is the velocity with which shot is fired......you can say initial velocity in newtons equation...
    The third part is crucial...You have to be careful...since initial velocity was in horizontal direction...and gravity acts vertically...so initial velocity will be unaffected....vertical velocity you can easliy calculate using newton's equation....(assuming initial vertical velocity is zero)...then find the net
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