Question

A 132 lb. (60 kg) person does a standing vertical jump. Starting from rest, she jumps vertically and reaches 70 cm (= 0.7m). The push-off phase takes 0.5 seconds. After being airborne, the landing takes 0.5 seconds before she comes to rest.

e. What is her vertical velocity when she becomes airborne? When she lands? f. How much mechanical work did the leg muscles produce during the push-off?

g. How much power did the leg muscles generate during the push off?

Answer #1

**e.Solution:**

mass of the person , m= 60 kg

let her vertical velocity when she becomes airborne = v

so her kinetic energy = 1/2 mv^{2}

from conservation of energy , initial kinetic energy = potential energy at height h=70 cm(0.7 m)

so potential energy at height ,h= 0.7 m is mgh where g is
acceleration due to gravity = 9.8 m/s^{2}

so from conservation of energy , mgh = 1/2 mv^{2}

or ,

so the velocity when she becomes airborne is 3.704 m/s

as the total energy is conserved , her velocity when she lands = 3.704m/s

**Answer: 3.704 m/s**

**f. Solution**:

Mechanical work done during push off by the legs = kinetic energy just after push off = potential energy at height 70 cm = mgh = 60 X 9.8 X 0.7 J

= **411.6 J**

Answer: **411.6 J**

**g. Solution :**

power generated during push off = work done/ time = 411.6/0.5 =
**823.2 J/s or Watt**

**Answer: 823.2 J/s or Watt**

Thank You.

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