Physics

A stone is thrown vertically upward with a speed of 18.0 m/s.
(a)How fast is it moving when it reaches a height of 11.0 m?
(b)How long is required to reach this height ?
(c)Why are there two answers to (b)?

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  1. first c
    It passes 11 meters on the way up and on the way down.

    now the calculations

    I guess it started at h = 0 (we are lying on our backs in a shallow hole throwing this upward )

    Vi = 18
    a = -9.81 m/s^2 = g

    initial Ke = (1/2) m v^2 = 162 m
    loss of Ke at 11 meters = m g h
    = 108 m
    remaining Ke at 11 meters = (162-108)m = 54 m
    so
    (1/2) m v^2 = 54 m
    v = 10.4 m/s on the way up, -10.4 m/s on the way down by symmetry(part a)

    part b
    v = Vi - 9.81 t
    10.4 = 18 - 9.81 t
    t = .775 seconds on the way up
    or
    -10.4 = 18 - 9.81 t
    - 28.4 = -9.81 t
    t = 2.9 seconds on the way down

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