# physics

A diver springs upward with an initial speed of 1.66 m/s from a 3.0-m board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = -3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.] (b) What is the highest point he reaches above the water?

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1. a. V^2 = Vo^2 + 2g*h
h(up) = (V^2-Vo^2)/2g.
h(up) = (0-2.76)/-19.6 = 0.141 m.

V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6(0.141+3) = 61.6
V = 7.85 m/s.

b. 0.141 m.

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2. CORRECTION:
b. hmax = 3 + 0.141 = 3.141 m.

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