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 1.75 m and cross the bar with a speed of 0.63 m/s?
energy at top=mgh+1/2 m vattop^2
energy at bottom:1/2 m vlaunch^2
set them equal, and solve for vlaunch
gfh
To determine the minimum speed with which the athlete must leave the ground in order to lift his center of mass to a certain height, we can use the principle of conservation of energy.
The initial kinetic energy of the athlete is converted into gravitational potential energy when he reaches the maximum height.
The formula for kinetic energy is given by:
KE = 0.5 * m * v^2
Where:
KE is the kinetic energy
m is the mass of the athlete
v is the velocity of the athlete
The formula for gravitational potential energy is given by:
PE = m * g * h
Where:
PE is the gravitational potential energy
m is the mass of the athlete
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height achieved
Since energy is conserved, the initial kinetic energy will be equal to the final gravitational potential energy:
0.5 * m * (initial velocity)^2 = m * g * h
Simplifying the equation:
0.5 * (initial velocity)^2 = g * h
Solving for the initial velocity:
(initial velocity)^2 = 2 * g * h
initial velocity = √(2 * g * h)
Now, plug in the given values:
g = 9.8 m/s^2 (acceleration due to gravity)
h = 1.75 m (height achieved)
initial velocity = √(2 * 9.8 * 1.75)
Calculating:
initial velocity ≈ 6.27 m/s
Therefore, the athlete must leave the ground with a minimum speed of approximately 6.27 m/s in order to lift his center of mass 1.75 m and cross the bar with a speed of 0.63 m/s.