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Mechanics questions

  1. Jan 13, 2008 #1
    Hello, i have a few questions i am doing and am wondering if someone would be kind enough to check i am doing them correctly:rolleyes:

    First question is..

    A shot is fired horizontally at a target 20m away and it is notice3d that the bullet hits a point 7.5cm below the target aimed at. What was the velocity of the bullet when it left the pistol?

    I get vertically
    u=0ms^1 a=9.8ms^2 s=0.075m need t
    so i used s=ut+0.5at^2
    t^2=0.075/4.9ms^2 so t =sqrt of 0.015 so t = 0.122seconds

    i the put this value int0 s=ut+0.5at^2
    so horizontally 20m=ux0.122+0.5(9.8ms^2x0.122^2)
    =20=ux0.122+0.073 so 19.937/0.122 =u so u = 163ms^1

    so the velocity on leaving the pistol was 163ms^1

    second question.

    A shot is fired with a velocity of 100ms^1 at an angle of 25degrees above a horizontal plane.Find...a) the time of flight on the horizontal plane b) the range on the horizontal plane. c)the greatest heiht the shot attains above the horizontal plane. d)the velocity of the shot (magnitude and direction)8.0seconds after it was fired.

    for a i get 8.73 seconds b) i get 790metres c)i get 91m and d) i get 120ms^1 at 25degree to the hoizontal.

    i am having a little bit of a job getting my head around this topic so if thes are not correct can someone help me out with what im doing wrong?

    Kind regards,
  2. jcsd
  3. Jan 13, 2008 #2

    Doc Al

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    Staff: Mentor

    So far, so good. Just be careful about how you round off your calculations. (Don't round off until the end.)

    Is the horizontal motion accelerated? (Which way does gravity act?)

    Don't just give your answers, show how you got them.
  4. Jan 13, 2008 #3
    hello, well the horizontal velocity is constant and g act downwards so only effects the vertical motion? i take it 163ms^1 is wrong.

    sorry for not putting my working for the second question. Here it is

    for question 2..

    V(horiz) = cos25degree x 100ms^1 so V(horiz) =90.6307787ms^1
    V(vert)= sin25degrees x 100ms^1 so V(vert)=42.3ms^1

    so taking the motion at max height
    A=g=9.8ms^2 need t

    so used V=u+AT
    0=42.26182617 +9.8 x t
    so t=42.26182617/9.8 = 4.32 seconds x 2 = 8.73seconds

    b) i just did range = horizontal velocity x flight time
    =90.6307787x8.73= 790m (3sf)

    c) t=4.32seconds u=42.26182617ms^1 v=0ms^1 s=h
    so = 42.26182617x4.32+0.5(-9.8x4.32^2)
    h = 91.1253291 metres

    d)at 8.0 sec g is acting at 78.4ms^2 downwards horizontal = constant 90.6307787ms^1
    so resultant velocity = v^2 = 78.4^2 + 90.6307787ms^1^2
    = 6146.56+8213.938048=v^2
    V=sqrt of 14360.49805 = 119.8352955ms^1

    to find the angle i used tan theta = 42.26182617/90.6307787 =tan-1 o.466307658=

    Hope you can help further, thankyou
    Chris :smile:
  5. Jan 13, 2008 #4

    Doc Al

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    Staff: Mentor

    Yes, it's wrong.

    Good. Except for that last step (x 2). :wink:

    Good. But correct the error propagated from part A.


    Redo this. What's the vertical component of velocity at 8 seconds? (g is always 9.8 m/s^2 downward.)
  6. Jan 13, 2008 #5
    sorry didn't mean to put the g=78.4 bit that would be the vertical velocity at 8.0sec and the horizontal is 90.6... so whatdo i do with that i thought i would find the resultant and then find the angle using trig? on the time should i have just left a 4.32sec? i thought it would be timesd by 2 as it is only halfway through the motion?

    Thankys for your help,
  7. Jan 13, 2008 #6
    okat 4.23 sec the vert velocity is 0 as the object is at max height so at 8 sec i work out that the vert velocity = 36.946ms^1 (3.77 secs after attaing max height?)
    is this correct?

  8. Jan 13, 2008 #7

    Doc Al

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    Staff: Mentor

    It's easier than all that. The vertical component of velocity is given by:
    [tex]v = v_0 - gt[/tex]

    Just plug in the numbers:
    [tex]v = 100\sin25 - (9.8)(8)[/tex]
  9. Jan 13, 2008 #8
    oh ok what is the v0 bit in that equation horizontal velocity? and for question 2 part a you said i was wrong to multiply the time of 4.23 seconds by two, how come? i thought this would be the max height reached so i thought i ha to multiply by 2 to get the total time of flight. Could you explain this for me as they did the same in my textbook?

  10. Jan 13, 2008 #9

    Doc Al

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    Staff: Mentor

    In my last post, I used v0 to stand for initial velocity in the vertical direction.

    It wasn't wrong to multiply by two, you just made an error when you did. 4.32 x 2 = 8.64 (not 8.73 like you wrote).
  11. Jan 14, 2008 #10
    Hello, i have just had a quick go at the first question i posted and is the answer 221.6ms-^1. I just did 300ms-^1sin90-g x t is this correct?

  12. Jan 14, 2008 #11

    Doc Al

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    Staff: Mentor

    Do you mean the first problem in your first post? If so, this isn't even close. (Your initial answer was very close, but your method was off a bit.)
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