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Math Help - physics question; how to find the velocity

  1. #1
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    physics question; how to find the velocity

    given horizontal distance=68 m
    angle with respect to horizontal, 35.8
    and time object spends in the air is 3.0 seconds.

    thanks.
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  2. #2
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    Hello, beetz!


    Given: horizontal distance = 68 m
    angle with respect to horizontal: 35.8
    time object spends in the air: 3.0 seconds.

    Find the initial velocity.

    We're expected to know this formula: . h \;=\;h_o + (v_o\sin\theta)t - 4.9t^2

    . . where: . \begin{array}{ccc}h_o & = & \text{initial height} \\<br />
v_o & = & \text{initial velocity} \\<br />
\theta & = & \text{angle of elevation} \\<br />
h & = & \text{height of object} \\<br />
 & & \text{at time }t<br />
 \end{array}


    We are given: . h_o = 0,\;\theta = 35.8^o

    . . Hence, we have: . h \;=\;(v_o\sin35.8^o)t - 4.9t^2

    The object is on the ground (h = 0) when t = 0 and t = 3.

    So we have: . (v_o\sin35.8^o)3 - 4.9(3^2) \;=\;0

    Therefore: . v_o \;=\;\frac{4.9(3^2)}{3\sin35.8^o} \;\approx\;\boxed{25.13\text{ m/sec}}

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  3. #3
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    thank you so much.
    the equation in my notes is a bunch of mumbo jumbo.
    though, almost everything in physics seems like a bunch of mumbo jumbo to me.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by beetz View Post
    given horizontal distance=68 m
    angle with respect to horizontal, 35.8
    and time object spends in the air is 3.0 seconds.

    thanks.
    You need to provide more detail otherwise we are just guessing at your question

    RonL
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