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Math Help - Brain Fart :/

  1. #1
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    Brain Fart :/

    a stone is tossed doown with a speed of 8 m/ s from the edge of a cliff 63 m high. how long will it take to hit the foot of the cliff?
    a= -9.8 m/s ^2
    vi= -8 m/s
    s= 63m

    63m= - 8 m/s(t) + 1/2(9.8m/s^2)(t)^2

    does this look right ?

    and if so how do i solve for t?

    Just confused can someone help me :/
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  2. #2
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    Quote Originally Posted by tnkfub
    a stone is tossed doown with a speed of 8 m/ s from the edge of a cliff 63 m high. how long will it take to hit the foot of the cliff?
    a= -9.8 m/s ^2
    vi= -8 m/s
    s= 63m

    63m= - 8 m/s(t) + 1/2(9.8m/s^2)(t)^2

    does this look right ?
    and if so how do i solve for t?
    Just confused can someone help me :/
    Hello,

    the stone starts at a height of 63 m. Then it is loosing height by being tossed and by falling, until it hits the ground, that means the height is zero(0).

    0= 63- 8\frac{m}{s} \cdot t- \frac{1}{2} \cdot 9.81 \frac{m}{s^2} \cdot t^2

    This is a quadratic equation in t. I suppose that you know how to solve a quadratic equation.

    You'll get x_1 \approx 2.86 s\ \vee \ x_2=-4.49 s

    The negative solution isn't very realistic.

    In comparison: If the stone were only fallen down, it had taken 3.58 s to come down. So the tossing gave the extra kick.

    Greetings

    EB
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tnkfub
    a stone is tossed doown with a speed of 8 m/ s from the edge of a cliff 63 m high. how long will it take to hit the foot of the cliff?
    a= -9.8 m/s ^2
    vi= -8 m/s
    s= 63m

    63m= - 8 m/s(t) + 1/2(9.8m/s^2)(t)^2

    does this look right ?

    and if so how do i solve for t?

    Just confused can someone help me :/
    Earboth didn't leave much to comment on, but I want to make a suggestion, considering the formula you came up with. It might seem like over-kill, but it's good practice to always draw a diagram and select a positive direction.

    For example, in your formula:
    a= -9.8 m/s ^2
    vi= -8 m/s
    s= 63m

    63m= - 8 m/s(t) + 1/2(9.8m/s^2)(t)^2
    You have apparently chosen the positive direction to be upward (that's why a and v are negative). s, however, also needs a direction...it's the overall displacement. Notice that the object ends up lower than it started, so s also should be negative. And when you put "a" into your equation, you switched the sign.

    Probably the biggest problem for an intro Physics student is getting the signs right. Even a quick diagram can fix these problems.

    -Dan
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