A tennis ball is dropped from 1.66 m above the
ground. It rebounds to a height of 0.959 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s^2. (Letdown be negative.)
Answer in units of m/s
(1/2) m v^2 = m g h
so
v = sqrt (2 g h) = sqrt (2*9.8*1.66)
down so negative
If it then asks for the speed headed back up do it the same way using 0.959 for height. Now positive because headed up.
To find the velocity with which the tennis ball hits the ground, we can use the principle of conservation of energy.
The total mechanical energy of the ball remains constant throughout its motion, so we can equate the potential energy at the initial height to the kinetic energy at the final height.
The potential energy (PE) is given by the formula: PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height.
The kinetic energy (KE) is given by the formula: KE = (1/2)mv^2, where v is the velocity of the ball.
At the initial height, the PE is equal to the potential energy at the final height since it rises to that height.
So, mgh = (1/2)mv^2
We can cancel out the mass (m) from both sides of the equation:
gh = (1/2)v^2
Now, solve for v:
v^2 = 2gh
Taking the square root of both sides:
v = √(2gh)
Given:
Height (h) = 1.66 m
Acceleration due to gravity (g) = 9.8 m/s²
Substituting these values into the equation:
v = √(2 * 9.8 * 1.66)
Calculating the value:
v = √(32.392) ≈ 5.69 m/s
Therefore, the velocity with which the tennis ball hits the ground is approximately 5.69 m/s.
To find the velocity with which the tennis ball hits the ground, we can use the concept of energy conservation. We know that the potential energy of the ball at the initial height is equal to the kinetic energy of the ball just before it hits the ground.
The potential energy of an object at a height h is given by the equation:
PE = mgh
where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
The kinetic energy of an object with mass m and velocity v is given by the equation:
KE = (1/2)mv^2
Since the potential energy at the initial height is equal to the kinetic energy just before hitting the ground, we can set up the following equation:
mgh = (1/2)mv^2
Simplifying the equation, we can cancel out the mass from both sides:
gh = (1/2)v^2
Next, we can solve for v by isolating it on one side of the equation:
v^2 = 2gh
v = √(2gh)
Now, we can substitute the given values into the equation:
g = 9.8 m/s^2
h = 1.66 m
v = √(2 * 9.8 m/s^2 * 1.66 m)
v ≈ 6.12 m/s
Therefore, the velocity with which the tennis ball hits the ground is approximately 6.12 m/s.