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While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 7.61 m/s. The stone subsequently falls to the ground, which is 18.1 m below the point where the stone leaves his hand. At what speed does the stone impact the ground? Ignore air resistance and use g=9.81 m/s^2 for the acceleration due to gravity.

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lcmendozaf

Answer:

V=20.35m/s

Explanation:

First of all we need to calculate the time it took the stone to reach ground level. We know that ΔY = -18.1m; Vo = 7.61m/s; g=9.81m/s2

[tex]\Delta Y = Vo*t - g*\frac{t^{2}}{2}[/tex]

[tex]-18.1=7.61*t-\frac{9.81*t^{2}}{2}[/tex]  Solving for t, we get:

t1 = -1.3s    t2 = 2.85s    We discard the negative solution and use the positive one. With this value we calculate the final velocity as:

Vf = Vo - g*t = 7.61 - 9.81 * 2.85 = -20.35 m/s

The speed is the module of the velocity, so:

V = 20.35 m/s

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