A 45.0 g golf ball is released 2.50 m above the ground. The velocity of the golf ball when it hits the ground is
what is the answer for this questions
A 45.0 g golf ball is released 2.50 m above the ground. Calculate the the velocity of the golf ball when it hits the ground in km/h.
45 g has nothing to do with this because m cancels
(1/2) m v^2 = m g h
v^2 = 2 g h
v = sqrt (2 g h) [ worth remembering ]
To find the velocity of the golf ball when it hits the ground, we can use the principle of conservation of energy. The initial potential energy of the golf ball is converted into kinetic energy as it falls.
First, let's calculate the potential energy of the golf ball when it is 2.50 m above the ground. The potential energy is given by the formula:
Potential Energy = mass * gravity * height
where mass is given as 45.0 g, gravity is approximately 9.8 m/s^2, and height is 2.50 m.
Converting the mass to kilograms:
mass = 45.0 g = 0.045 kg
Now, we can calculate the potential energy:
Potential Energy = 0.045 kg * 9.8 m/s^2 * 2.50 m
Next, we can equate the potential energy to the kinetic energy when the ball hits the ground. The formula for kinetic energy is:
Kinetic Energy = 0.5 * mass * velocity^2
Since the ball starts from rest, its initial kinetic energy is 0. Therefore, the final potential energy is equal to the final kinetic energy at the moment it hits the ground.
Setting the two equal, we have:
Potential Energy = Kinetic Energy
0.045 kg * 9.8 m/s^2 * 2.50 m = 0.5 * mass * velocity^2
Simplifying:
0.045 kg * 9.8 m/s^2 * 2.50 m = 0.5 * 0.045 kg * velocity^2
Now, we solve for velocity:
velocity^2 = (0.045 kg * 9.8 m/s^2 * 2.50 m) / (0.5 * 0.045 kg)
velocity^2 = 9.8 m/s^2 * 2.50 m
Taking the square root of both sides:
velocity = sqrt(9.8 m/s^2 * 2.50 m)
Calculating:
velocity ≈ 7.96 m/s
Therefore, the velocity of the golf ball when it hits the ground is approximately 7.96 m/s.
To find the velocity of the golf ball when it hits the ground, you can use the principle of conservation of energy.
Given:
- Mass of the golf ball (m) = 45.0 g = 0.045 kg
- Height from which the ball is released (h) = 2.50 m
The potential energy (PE) of an object of mass (m) at a height (h) above the ground is given by the equation:
PE = m * g * h
Where:
- m = mass of the object
- g = acceleration due to gravity (approximately 9.8 m/s^2)
- h = height above the ground
The potential energy at the topmost point (when the ball is released) will be converted to kinetic energy at the bottommost point (when it hits the ground) since there is no loss of energy due to friction or air resistance.
The kinetic energy (KE) of an object of mass (m) and velocity (v) is given by the equation:
KE = (1/2) * m * v^2
Since energy is conserved, we can equate the potential energy to the kinetic energy:
m * g * h = (1/2) * m * v^2
Now, we can solve for the velocity (v):
v^2 = 2 * g * h
v = sqrt(2 * g * h)
Substituting the given values:
v = sqrt(2 * 9.8 m/s^2 * 2.50 m)
v ≈ sqrt(49 m^2/s^2)
v ≈ 7 m/s
Therefore, the velocity of the golf ball when it hits the ground is approximately 7 m/s.