A tennis ball is dropped from 1.76 m above the
ground. It rebounds to a height of 0.928 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s
2
I answered a virtually identical question recently, just a different drop height.
v = sqrt (2 g h)
to make it simple but for the two ways to get there see:
https://www.jiskha.com/questions/1802335/A-tennis-ball-is-dropped-from-1-93-m-above-the-ground-It-rebounds-to-a-height-of
By the way, the answer does not depend on the rebound. You hit at a certain speed v = sqrt (2 g h) ignoring air friction. Your rebound height depends on your elasticity (coef of restitution) and the ground consistency (not squishy)
To calculate the velocity at which the tennis ball hits the ground, we can use the concept of conservation of mechanical energy.
The initial mechanical energy of the ball when it is dropped is equal to its final mechanical energy when it rebounds.
The initial mechanical energy of the ball is given by the potential energy at the height it is dropped from, which is the product of its mass (m), the acceleration due to gravity (g), and the height (h):
Initial potential energy = mgh
The final mechanical energy of the ball is given by its kinetic energy just before it hits the ground:
Final kinetic energy = (1/2)mv^2
Since there is no loss in mechanical energy during the rebound, the initial potential energy is equal to the final kinetic energy.
So, mgh = (1/2)mv^2
We can cancel the mass (m) and solve for the velocity (v):
gh = (1/2)v^2
v^2 = 2gh
Substituting the given values:
v^2 = 2 * 9.8 m/s^2 * 1.76 m
v^2 = 34.536 m^2/s^2
Taking the square root of both sides:
v = √(34.536 m^2/s^2)
v ≈ 5.88 m/s
Therefore, the tennis ball hits the ground with a velocity of approximately 5.88 m/s.
To find the velocity with which the tennis ball hits the ground, we can use the law of conservation of energy. According to this law, the total mechanical energy of an object remains constant if no external forces are acting on it.
The mechanical energy of an object can be divided into two parts: kinetic energy (KE) and potential energy (PE). Kinetic energy is the energy of motion, and potential energy is the energy associated with an object's position.
At the topmost point of its path, the tennis ball has only potential energy, given by the equation PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height.
At the bottommost point of its path (when it hits the ground), the tennis ball has only kinetic energy, given by the equation KE = 1/2mv^2, where v is the velocity of the ball.
Since mechanical energy is conserved, the potential energy at the topmost point is equal to the kinetic energy at the bottommost point:
mgh = 1/2mv^2
We can cancel out the mass (m) on both sides of the equation:
gh = 1/2v^2
Rearranging the equation, we get:
v^2 = 2gh
Taking the square root of both sides of the equation, we can find the velocity (v):
v = √(2gh)
Now, we can substitute the given values to find the velocity at which the tennis ball hits the ground:
h = 1.76 m
g = 9.8 m/s^2
Plugging these values into the equation, we get:
v = √(2 * 9.8 * 1.76) = √34.496 = 5.87 m/s
Therefore, the velocity with which the tennis ball hits the ground is approximately 5.87 m/s.