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Math Help - A ball is thrown

  1. #1
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    A ball is thrown

    A ball is thrown stright up with the speed 25m/s

    how high does the ball go and at what time is it at half its height?

    I think am suppose to use somthing like s=v*t but i dont know how to find the time or the s so i can solve for the other. I also know gravity should play a part in the speed
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JBswe View Post
    A ball is thrown stright up with the speed 25m/s

    how high does the ball go and at what time is it at half its height?

    I think am suppose to use somthing like s=v*t but i dont know how to find the time or the s so i can solve for the other. I also know gravity should play a part in the speed
    i believe you are referring to suvat equations, those can work, but i never remember them. let's start from scratch

    the acceleration due to gravity is -9.81 m/s

    thus we have:

    a(t) = -9.81 with the initial condition, v(0) = 25 m/s and s(0) = 0

    now, let's find v(t) and s(t)

    (by the way s(t) is the position function, v(t) is the velocity function and a(t) is the acceleration function)

    so, v(t) = \int a(t)~dt = \int -9.81~dt = -9.81t + C

    since v(0) = 25, we have:

    v(0) = 25 = -9.81(0) + C \implies C = 25

    thus, v(t) = -9.81t + 25

    we obtain the max height when v(t) = 0

    that is, when t \approx 2.548

    plug that into the position function:

    s(t) = \int v(t)~dt = \int (-9.81t + 25)~dt = -4.9t^2 + C

    since s(0) = 0, we have that C = 0 here.


    since we obtain the max height at t = 2.548, the max height is, s(2.548)

    i think you can handle the rest


    ...by the way? have you done calculus? if not, the suvat equations are the way to go, forget everything i said and see here (note that these equations assume all initials are zero.
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