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Math Help - [SOLVED] Initial velocity using Calculus not physics??

  1. #1
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    [SOLVED] Initial velocity using Calculus not physics??

    I have the equation x=ct-4.9t^2 where x=height, t=time for a body projected vertically up from the ground.

    I want to find the initial velocity if the body just reaches a height of 250m using calculus (I already found the answer with Physics-see below).

    The maths
    x=ct-4.9t^2
    x`=c-9.8t, c=9.8t
    x``=-9.8

    Can someone help?


    s=(final velocity^2-initial velocity^2)/2A
    250= (0^2-x^2)/2(-9.8)
    X=70m/s
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  2. #2
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    Quote Originally Posted by Neverquit View Post
    I have the equation x=ct-4.9t^2 where x=height, t=time for a body projected vertically up from the ground.

    I want to find the initial velocity if the body just reaches a height of 250m using calculus (I already found the answer with Physics-see below).

    The maths
    x=ct-4.9t^2
    x`=c-9.8t, c=9.8t
    x``=-9.8

    Can someone help?


    s=(final velocity^2-initial velocity^2)/2A
    250= (0^2-x^2)/2(-9.8)
    X=70m/s
    "Using physics rather than calculus" strikes me as an odd expression since the physics equations are derived from calculus, but I understand your meaning.

    At the peak of its trajectory, dx/dt = 0, so we set dx/dt = c - 9.8t = 0

    Solve for t, to get t = c/9.8

    This is the time at which x_max is reached

    plug back into equation for x(t)

    x_max = ct_max - 4.9t_max^2

    250 = c(c/9.8) - 4.9(c/9.8)^2

    ...

    c = 70 m/s
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