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18 April, 02:10

A small block is placed at height h on a frictionless 30 degree ramp. Upon being released the block slides down the ramp and then falls 1.0m to the floor. A small hole is located 1.0 m from the end of the ramp. From what height h should the block be released in order to land in the hole?

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  1. 18 April, 03:34
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    After leaving the plane, the block will have an unknown speed (S),

    which can be broken into x, y components.

    The x, y kinematics are: x - 1

    x0 - 0 V - ? V0 - Scos (-30)

    a - 0

    t - t

    y - 0

    y0 - 1

    V - ?

    V0 - Ssin (-30)

    a - - 9.8

    t - t

    We then use x=x0+v0t+.5at^2

    in the x case: 1=0+Scos (-30) +.5 (0) t^2

    Solving for t gives t=1 / Scos (-30)

    in the y case,

    with t-substitution:

    0=1+Ssin (-30) * 1/Scos (-30) +.5 (-9.8) (1/Scos (-30)) ^-2

    In the middle velocity term, S cancels out. Multiplying all known numbers as well as squaring the third term gives:

    0=1-.5774-6.5333/S^2

    Solving for S = S = 3.9319 m/s

    Now with a mark on final ramp speed, we can now make a 3rd kinematics equation. The acceleration will be altered from gravity:

    Slide force = 9.8*sin (30) = 4.9 m/s^2.

    x - ?

    x0 - 0

    V - 3.9319

    V0 - 0

    a - 4.9

    t - ?

    So the equation we use is V2 = V02+2a (x-x0). 3.93192=0+2*4.9 * (x-0)

    Solving for x gives x=1.5775 m up the ramp.

    So we now look for the y component of the ramp length:

    1.5775*sin (30) =.78875 m 'high' on the ramp.
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