A 78 kg volleyball player jumps and leaves the ground with a vertical
velocity of 3.5 m/s. By thinking about the conservation of energy in a
freefalling object, work out how high their centre of mass would reach at
the top of their jump and present your answer to the nearest centimetre.
See for example:
http://en.wikipedia.org/wiki/Coulomb's_law
Oops,wrong post.
m=78kg
v=3.5 m/s
With energy conserved,
PE=KE
mgh=(1/2)mv²
h=v²/(2g)
=3.5²/(2*9.8) m
Convert result to nearest centimetre.
To find the height reached by the volleyball player at the top of their jump, we can use the principle of conservation of energy.
The total mechanical energy of an object is the sum of its kinetic energy and potential energy. At the top of their jump, when the player momentarily comes to rest, all of their initial kinetic energy is converted into potential energy.
The kinetic energy of an object can be calculated using the formula:
Kinetic Energy = 1/2 * mass * velocity^2
Given that the mass of the player is 78 kg and the vertical velocity is 3.5 m/s, we can find the kinetic energy:
Kinetic Energy = 1/2 * 78 kg * (3.5 m/s)^2
Next, we can equate this kinetic energy to the potential energy at the top of the jump. The potential energy of an object near the surface of the Earth is given by the formula:
Potential Energy = mass * gravitational acceleration * height
The gravitational acceleration near the surface of the Earth is approximately 9.8 m/s^2.
Setting the kinetic energy equal to the potential energy, we have:
1/2 * 78 kg * (3.5 m/s)^2 = 78 kg * 9.8 m/s^2 * height
Now we can solve for height:
height = (1/2 * 78 kg * (3.5 m/s)^2) / (78 kg * 9.8 m/s^2)
Simplifying the equation gives us:
height = (0.5 * 3.5^2) / 9.8
height ≈ 0.61 meters
Therefore, the center of mass of the volleyball player reaches approximately 61 centimeters at the top of their jump.