Answer:
the velocity with which the tennis ball leaves the ground is approximately 5.996 meters per second
Step-by-step explanation:
The potential energy (PE) of an object at a certain height h above the ground in a gravitational field can be calculated using the formula:
PE = m * g * h
Where:
PE is the potential energy
m is the mass of the object (not given, but it will cancel out in our calculation)
g is the acceleration due to gravity (9.8 m/s²)
h is the height above the ground
Initial potential energy when dropped:
PE_initial = m * g * h_initial = 9.8 m/s² * 1.83 m = 17.934 J
Final potential energy when at the highest point after rebounding:
PE_final = m * g * h_final = 9.8 m/s² * 0.872 m = 8.5616 J
At the highest point, all the initial potential energy is converted into kinetic energy (KE) since the speed momentarily becomes zero. So:
PE_initial = KE_final
Therefore, the kinetic energy at the highest point is also 17.934 J.
The kinetic energy of an object can be calculated using the formula:
KE = 0.5 * m * v^2
Where:
KE is the kinetic energy
m is the mass of the object (not given, but it will cancel out in our calculation)
v is the velocity
Substituting the known value for KE_final:
17.934 J = 0.5 * v^2
Now, solve for v:
v^2 = (2 * 17.934 J) / 1
v^2 = 35.868 J
v = sqrt(35.868 J)
v ≈ 5.996 m/s
So the velocity with which the tennis ball leaves the ground is approximately 5.996 meters per second