In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what 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?
To determine the minimum speed the athlete must leave the ground with, we can use the principle of conservation of energy. According to this principle, the total mechanical energy (kinetic energy + potential energy) remains constant in the absence of external forces like friction or air resistance.
Initially, when the athlete leaves the ground, all of their energy is in the form of kinetic energy. As they rise, this kinetic energy is gradually converted into gravitational potential energy, given by the equation:
mgh = (1/2)mv^2
where m is the mass of the athlete, g is the acceleration due to gravity (approximately 9.8 m/s^2), h is the change in height (1.65 m), and v is the final velocity (0.75 m/s).
To find the minimum speed the athlete must leave the ground with, we need to determine the initial velocity (u) at the start of the jump. We can rearrange the equation to solve for u:
(1/2)mu^2 = mgh - (1/2)mv^2
Simplifying further:
u^2 = 2gh - v^2
Substituting the known values:
u^2 = 2 * 9.8 * 1.65 - 0.75^2
u^2 = 32.34 - 0.5625
u^2 ≈ 31.78
Taking the square root of both sides:
u ≈ √31.78
u ≈ 5.64 m/s
Therefore, the minimum speed the athlete must leave the ground with to lift their center of mass 1.65 m and cross the bar with a speed of 0.75 m/s is approximately 5.64 m/s.