v = 2 a Δ y , were a = (7/3) g and Δ y = 0.600 m. This causes the ball to reach a maximum height h max (measured above the water surface) given by h max = v 2 /2 g (see Eq. 2-16 again). Thus, h max = (7/3) Δ y = 1.40 m. 44. Due to the buoyant force, the ball accelerates upward (while in the water) at rate a given by Newton’s second law: ρ water Vg – ball Vg = ball Va ¡ ball = water (1 + “ a ”) where – for simplicity – we are using in that last expression an acceleration “
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