When a 65 kg skydiver jumps from a plane, her speed steadily increases until air resistance provides a force that balances the force due to free fall. How fast is the
skydiver falling if her kinetic energy at the moment is 704 x 10- J?

Correct value of kinetic energy is 704 × 10³ J

v = 147.18 m/s

Explanation:

At any point/moment during the fall, the sky diver will possess kinetic energy with the formula;

KE = ½mv²

Where;

m is mass

v is velocity

We are given;

m = 65 kg

KE = 704 × 10³ J

Thus;

704 × 10³ = ½ × 65 × v²

v² = (2 × 704 × 10³)/65

v² = 21661.53846153846

v = √21661.53846

v = 147.18 m/s

the answer would be 32*f