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Math Help - motion in one-d

  1. #1
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    motion in one-d

    ball thrown up with v=10m/s from a building of height 100. find max height and time taken to reach this max.
    i said a=-g

    integrate to get v=-gt+C but initially v=10 so v=-gt+10

    at max height v=0 so t=10/g

    now i integrate again to get s=-(gt^2)/2+10+K but initially s=100 so K=100

    so at t=10/g, i get s= 5.1

    this answer is correct but they use

     v_{0}^2-2g \Delta h=v

    can anyone show me how they got this equation?

    For part b) they want the time taken to go from h=50 to h=0.

    for this i used the formula above for s and solved for t wneh h-50 and 0 to get the correct answer.

    in the solutions they use
     v_{0}^2-2g \Delta h=v

    and if v2 is velocity at H=0 and v1 is velocity at H=50 then

    v2=v1-gt

    is the above obvious?im thinking,wrongly, if i start motion at v1 then f=ma gives me a=g so v=gt+K but initially v=v1 thus v=gt+v1 ??

    Can i get v2=v1-gt starting with f=ma?
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  2. #2
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    Quote Originally Posted by jiboom View Post
    ball thrown up with v=10m/s from a building of height 100. find max height and time taken to reach this max.
    i said a=-g

    integrate to get v=-gt+C but initially v=10 so v=-gt+10

    at max height v=0 so t=10/g

    now i integrate again to get s=-(gt^2)/2+10+K but initially s=100 so K=100
    This should be s= -gt^2/2+ 10t+ K.

    so at t=10/g, i get s= 5.1
    What? a ball is thrown upward from the top of a 100 m building and you assert that its highest point it 5.1 m??

    What value did you use for g? 9.8? -9.8(1)/2+ 10(1)+ 100= 105.1. If you are calculating "height above the building" then your "K" is 0.

    this answer is correct but they use

     v_{0}^2-2g \Delta h=v

    can anyone show me how they got this equation?
    Did they really set that equal to v? From your v= -gt+ v_0, t= (v_0- v)/g. Put that into s= \Delta h= (-g/2)t^2+ v_0t and you get \Delta h= -(g/2)(v_0^2- 2v_0v+ v^2)/g^2+ v_0(v_0- v)/g= -v_0^2/2g+ v_0v/g- v^2/2g+ v_0^2/g- v_0v/g = v_0^2/2g- v^2/2g . Multiply both sides by 2g to get 2g\Delta h= v_0^2- v^2 which is the same as v^2= v_0^2- 2g\Delta h= 0, not v.

    You can also interpret that in terms of energy. At velocity v, the kinetic energy is (1/2)mv^2 and at height \Delta h the potential energy is mg \delta h. Multiply both sides of v^2= v_0^2- 2g\Delta h by m/2 and you have (1/2)mv^2= (1/2)mv_0^2- mg\Delta h or (1/2)mv^2+ mg\Delta h= (1/2)mv_0^2- that the total energy, kinetic energy plus potential energy, at any point is equal to the kinetic energy at the start where the potential energy was 0.

    For part b) they want the time taken to go from h=50 to h=0.

    for this i used the formula above for s and solved for t wneh h-50 and 0 to get the correct answer.

    in the solutions they use
     v_{0}^2-2g \Delta h=v

    and if v2 is velocity at H=0 and v1 is velocity at H=50 then

    v2=v1-gt

    is the above obvious?im thinking,wrongly, if i start motion at v1 then f=ma gives me a=g so v=gt+K but initially v=v1 thus v=gt+v1 ??
    a= -g doesn't it? so that v= -gt+ v1?

    Can i get v2=v1-gt starting with f=ma?
    Yes, that's what I just said: v= -gt+ v1 is the same as v2= v1- gt.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post

    What? a ball is thrown upward from the top of a 100 m building and you assert that its highest point it 5.1 m??

    .
    bit rushed typing it in, i do realise i need 100+5.01 Im bad at this stuff but not that bad yet

    Quote Originally Posted by HallsofIvy View Post
    a= -g doesn't it? so that v= -gt+ v1?
    just to verify, this is -g because we initially throw it up ? If i just dropped it then iwould choose the direction so a=g?
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