A 72.0 kg swimmer jumps into the old swimming hole from a tree limb that is 3.95

Question

A 72.0 kg swimmer jumps into the old swimming hole from a tree limb that is 3.95 m above the water.

Use energy conservation to find his speed just as he hits the water if he just holds his nose and drops in.

Express your answer to three significant figures.

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Part B

Use energy conservation to find his speed just as he hits the water if he bravely jumps straight up (but just beyond the board!) at 2.60 m/s .

Express your answer to three significant figures.

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Part C

Use energy conservation to find his speed just as he hits the water if he manages to jump downward at 2.60 m/s .

Express your answer to three significant figures.

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Explanation / Answer

a)

m = mass of swimmer = 72 kg

h = height from which the swimmer jumps = 3.95 m

v = speed of swimmer just as it hits the water

using conservation of energy

kinetic energy gained = potential energy lost

(0.5) m v2 = mgh

v = sqrt(2gh)

v = sqrt(2 x 9.8 x 3.95)

v = 8.8 m/s

b)

vi = initial speed = 2.60 m/s

h = height from which the swimmer jumps = 3.95 m

vf = speed of swimmer just as it hits the water

using conservation of energy

kinetic energy at the top + potential energy at top = kinetic energy at bottom

(0.5) m vi2 + mgh = (0.5) m vf2

vi2 + 2gh = vf2

(2.60)2 + 2(9.8 x 3.95) = vf2

vf = 9.2 m/s

c)

vi = initial speed downward = 2.60 m/s

h = height from which the swimmer jumps = 3.95 m

vf = speed of swimmer just as it hits the water

using conservation of energy

kinetic energy at the top + potential energy at top = kinetic energy at bottom

(0.5) m vi2 + mgh = (0.5) m vf2

vi2 + 2gh = vf2

(2.60)2 + 2(9.8 x 3.95) = vf2

vf = 9.2 m/s

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