In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. Find the minimum speed must the athlete leave the ground in order to lift his center of mass 1.65 m and cross the bar with a speed of 0.75 m/s.

The answer to your question is v₁ = 5.74 m/s

Explanation:

Data

v₁ = ?

h₁ = 0 m

v₂ = 0.75 m/s

h₂ = 1.65 m

g = 9.81 m/s²

Formula

mgh₁ + 1/2mv₁² = mgh₂ + 1/2mv₂²

mass is not consider (if we factor mass, it is cancelled)

gh₁ + 1/2v₁² = gh₂ + 1/2v₂²

Substitution

(9.81) (0) + 1/2v₁² = (9.81) (1.65) + 1/2 (0.75) ²

Simplification

0 + 1/2v₁² = 16.19 + 0.28

Solve for v₁

1/2v₁² = 16.47

v₁² = 2 (16.47)

v₁² = 32.94

Result

v₁ = 5.74 m/s