Sam tosses a ball horizontally off a footbridge at 3.1 m/s. How much time passes after he releases it until its speed doubles?
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Expert's answer
2010-12-22T03:26:42-0500
The horizontal projection is constant while the vertical is changing according to the equation Vy=V0y+gt (we chose the downward direction as positive). Vx = 3.1 m/s V0y = 0; V0 = 3.1 m/s V = √(Vx2 + Vy2) = √(3.12 +100t2) V2/V02 = 4 (3.12 +100t2)/ 3.12 = 4 t =√(3.12*3 / 100) = 0.54 s
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