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Math Help - Falling object

  1. #1
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    Arrow Falling object

    Hi...

    Q) A stone is thrown vertically upwards with a speed of 16m/s from a point h metres above the ground. The stone hits the ground 4s later. Find the value of h.

    Using suvat equations, how is this done?

    Thanks alot
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Apex View Post
    Hi...

    Q) A stone is thrown vertically upwards with a speed of 16m/s from a point h metres above the ground. The stone hits the ground 4s later. Find the value of h.

    Using suvat equations, how is this done?

    Thanks alot
    the SUVAT equation we need here is:

    s = vt - \frac {at^2}{2} = ut + \frac {at^2}{2}

    where u is the initial velocity, v is the final velocity, s is the displacement or position, and a is the acceleration.

    However, since we did not start on the ground, s has to be modified by some constant. that is, we will use

    s = ut + \frac {at^2}{2} + C where C is the height we are when time, t, is zero.

    now u=16, and a = -9.8

    So, s = 16t + \frac {-9.8t^2}{2} + C

    \Rightarrow s = 16t - 4.9t^2 + C

    Now, after 4 seconds, the ball hits the ground, so it's height is zero.

    \Rightarrow s(4) = 0

    \Rightarrow 0 = 16(4) - 4.9(4)^2 + C

    \Rightarrow C = 14.4

    So, s = -4.9t^2 + 16t + 14.4

    So h = 14.4, since that is what s is when t = 0
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  3. #3
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    Thanks...I kept getting something like 13.06(my way). The solution in the paper returns the same answer as you, but it is done like this:

    s=ut + 0.5at^2

    and
    s= ?
    u= 16
    v= -
    a= -9.8
    t= 4

    What I don't understand, is why if it's done this way (using these values), it actually gives the value of h, as oppose to the distance from h, to where v=0, and then to the ground. i.e., the whole journey of the rock minus the distance h.

    Can you please explain how this method works?
    Thanks
    Apex
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Apex View Post
    Thanks...I kept getting something like 13.06(my way). The solution in the paper returns the same answer as you, but it is done like this:

    s=ut + 0.5at^2

    and
    s= ?
    u= 16
    v= -
    a= -9.8
    t= 4

    What I don't understand, is why if it's done this way (using these values), it actually gives the value of h, as oppose to the distance from h, to where v=0, and then to the ground. i.e., the whole journey of the rock minus the distance h.

    Can you please explain how this method works?
    Thanks
    Apex
    think of it this way. we can use h instead of C.

    Now we know that s = ut + 0.5at^2. however, this equation assumes that the displacement is zero when time is zero. (if you plug in t=0 we get 0). here that is not the case. when time is zero, we are at a distance h above the ground. so we must have:

    s = ut + 0.5at^2 + h

    Now if we plug in t=0, we get the distance h, which is what we want.

    Now we simply must find h. we know that s = 0 when t = 4, so we simply plug those values in to find h

    i'm not sure if i answered your question. did i?
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  5. #5
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    Thumbs up

    Oh right, I see what you mean.

    Thanks very much for your help

    br
    Apex
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