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25 August, 11:59

While standing on a bridge 15 m above the ground, you drop a stone from rest. When the stone has fallen 3.2 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant?

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  1. 25 August, 12:09
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    11.3 m/s

    Explanation:

    First, find the time it takes for the first stone to fall 3.2 m.

    Given:

    Δy = 3.2 m

    v₀ = 0 m/s

    a = 9.8 m/s²

    Find: t

    Δy = v₀ t + ½ at²

    (3.2 m) = (0 m/s) t + ½ (9.8 m/s²) t²

    t = 0.81 s

    Next, find the time for the first stone to land.

    Given:

    Δy = 15 m

    v₀ = 0 m/s

    a = 9.8 m/s²

    Find: t

    Δy = v₀ t + ½ at²

    (15 m) = (0 m/s) t + ½ (9.8 m/s²) t²

    t = 1.75 s

    The difference in time is 1.75 s - 0.81 s = 0.94 s. Find the initial velocity needed for the second stone to land after that amount of time.

    Given:

    Δy = 15 m

    a = 9.8 m/s²

    t = 0.94 s

    Find: v₀

    Δy = v₀ t + ½ at²

    (15 m) = v₀ (0.94 s) + ½ (9.8 m/s²) (0.94 s) ²

    v₀ = 11.3 m/s
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