A 48 kg diver jumps off a cliff (with a running start) into the ocean. The cliff

Question

A 48 kg diver jumps off a cliff (with a running start) into the ocean. The cliff is 50 m above the ocean below. Her coach, using a video of the dive, determines that at a point in flight when she has risen 0.7 m above the cliff, her speed (center of mass) is 0.5 m/s. Frictional effects such as drag are negligible. Formulate your solution using the diver and Earth's gravity field as a system. Gravity does not do work on this system. It's effects are captured in changes in potential energy. How much kinetic energy did she have at takeoff? What was her speed? How much kinetic energy will she have as she splashes into the ocean? What minimum amount of chemical energy needed to be consumed within the diver's body in order for her to walk to the cliff, from ocean level, and then take off (jump)? Explain how you know.

Explanation / Answer

a)

let assume the topmost point of cliff as the reference point .

h = height gained above the cliff = 0.7 m

V = speed at height "h' = 0.5 m/s

m = mass = 48 kg

Vo = speed at the cliff just after jump

Using conservation of energy

Kinetic energy at the Cliff = KE at height "h" + PE at height "h"

(0.5) m Vo2 = (0.5) m V2 + mgh

Vo2 = V2 + 2gh

Vo2 = 0.72 + 2 x 9.8 x 0.7

Vo = 3.77 m/s

b)

Assuming the surface of ocean as reference point

H = height of cliff = 50

V' = speed at the surface of ocean

using conservation of energy

KE at Cliff + PE at Cliff = KE at the surface of ocean

(0.5) m Vo2 + mgH = (0.5) m V'2

Vo2 + 2gH = V'2

3.772 + 2 (9.8) (50) = V'2

V' = 31.5 m/s

c)

Chemical energy = mgH + (0.5) m Vo2 = 48 (9.8) 50 + (0.5) (48) (3.77)2 = 23861.11 J

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