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11 September, 08:02

A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building 1.60 s after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground 7.00 s after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?

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  1. 11 September, 08:35
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    a) 15.68 m/s

    b) 130.34 m

    Explanation:

    Let's consider the origin of the coordinates at the ground, therefore, the equation of motion of the rock and its velocity are

    x (t) = h + vt - (1/2) gt^2

    v (t) = v - gt

    Where h is the height of the building and v is the initial velocity.

    The maximum height is reached when v=0, that is v (1.6s) = 0, and we know that x (7s) = 0

    Therefore

    0 = v (1.6s) = v - (9.8 m/s^2) (1.6s)

    v = 15.68 m/s

    and

    0 = x (7s) = h + (15.68 m/s) (7 s) - (1/2) (9.8 m/s^2) (49s^2)

    h = (1/2) (9.8 m/s^2) (49s^2) - (15.68 m/s) (7 s)

    h = 130.34 m
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