The rebounder in basketball has a vertical leap (that is, the vertical movement of a fixed point on their body) of 120 cm . What is his initial "launch" speed off the ground?
To determine the initial "launch" speed of the rebounder, we can use the concept of conservation of energy. The rebounder's initial potential energy when standing on the ground is converted into kinetic energy as they leave the ground.
The potential energy (PE) of an object at a certain height h is given by the equation:
PE = m * g * h
Where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
In this case, we need to convert the vertical leap of 120 cm to meters by dividing by 100:
h = 120 cm / 100 = 1.20 m
Now, we can calculate the potential energy of the rebounder at the highest point of the leap:
PE = m * g * h
The kinetic energy (KE) of an object is given by the equation:
KE = (1/2) * m * v^2
Where v is the velocity of the object.
At the highest point of the leap, the rebounder will have no potential energy left, so the initial potential energy is equal to the final kinetic energy:
PE = KE
m * g * h = (1/2) * m * v^2
Since the mass of the rebounder cancels out, we can solve for v:
g * h = (1/2) * v^2
2 * g * h = v^2
v^2 = 2 * g * h
v = √(2 * g * h)
Now, substitute the known values:
v = √(2 * 9.8 m/s² * 1.20 m)
v = √(23.52 m²/s²)
v ≈ 4.85 m/s
Therefore, the rebounder's initial "launch" speed off the ground is approximately 4.85 m/s.