A rubber ball of mass 0.12 kg is held at a height of 2.5 m above the ground, then released.

Calculate the velocity of the ball before it reaches the ground.

Change in potential energy = Kinetic Energy

m*g*h = (1/2)*m*(v^2)
9.8 * 2.5 * 2 = v^2
49 = v^2

v = 7 m/s

V^2 = Vo^2 + 2g*h = 0 + 19.6*2.5 = 49,

V = 7 m/s.

Well, you know what they say, "What goes up must come down... with style!" So let's calculate the velocity of this rubber ball as it gracefully descends to the ground.

We can start by using the equation for gravitational potential energy (PE) to find the initial potential energy of the ball when it is held at a height of 2.5 m:

PE = m * g * h

Where:
m = mass of the ball (0.12 kg)
g = acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
h = height (2.5 m)

Substituting the values in, we get:
PE = 0.12 kg * 9.8 m/s^2 * 2.5 m

Now, let's assume that all the potential energy is converted into kinetic energy (KE) when the ball reaches the ground. The kinetic energy formula is given by:

KE = (1/2) * m * v^2

Where:
m = mass of the ball (0.12 kg)
v = velocity of the ball

Since PE is equal to KE, we can equate the two equations and solve for v:

PE = KE

0.12 kg * 9.8 m/s^2 * 2.5 m = (1/2) * 0.12 kg * v^2

Now cancel out the terms and simplify:

2.94 = 0.06 * v^2

Divide both sides by 0.06:

v^2 = 2.94 / 0.06

v^2 ≈ 49

Taking the square root of both sides:

v ≈ √49

v ≈ 7 m/s

So, my friend, the rubber ball will be cruising at approximately 7 meters per second before it reaches the ground. Watch out below!

To calculate the velocity of the ball before it reaches the ground, you can use the concept of conservation of energy. The potential energy at the top is converted into kinetic energy as the ball falls.

Step 1: Determine the potential energy at the top.
The potential energy of an object of mass (m) and height (h) above the ground is given by the formula:
Potential energy = mass × acceleration due to gravity × height
In this case, the mass of the ball is 0.12 kg, and the height is 2.5 m. The acceleration due to gravity is approximately 9.8 m/s^2.
Potential energy = 0.12 kg × 9.8 m/s^2 × 2.5 m

Step 2: Calculate the potential energy.
Potential energy = 0.12 kg × 9.8 m/s^2 × 2.5 m = 2.94 Joules

Step 3: Calculate the velocity.
The potential energy is converted into kinetic energy at the bottom, where the velocity is maximum. The kinetic energy (KE) is given by the formula:
Kinetic energy = (1/2) × mass × velocity^2
Using this formula, we can solve for the velocity at the bottom of the fall.
2.94 Joules = (1/2) × 0.12 kg × velocity^2

Step 4: Solve for the velocity.
Multiply both sides of the equation by 2 and divide both sides by the mass to isolate the velocity squared.
2 × 2.94 Joules = 0.12 kg × velocity^2
5.88 Joules = 0.12 kg × velocity^2

Take the square root of both sides to solve for the velocity.
Velocity = √(5.88 Joules / 0.12 kg)
Velocity ≈ 7.00 m/s

Therefore, the velocity of the ball before it reaches the ground is approximately 7.00 m/s.

To calculate the velocity of the ball before it reaches the ground, we can use the law of conservation of energy.

The potential energy (PE) of the ball when it is at a height of 2.5 m can be calculated using the equation PE = mgh, where m is the mass of the ball (0.12 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height (2.5 m).

PE = (0.12 kg) * (9.8 m/s^2) * (2.5 m)
= 2.94 joules

According to the law of conservation of energy, the potential energy is converted into kinetic energy (KE) as the ball falls. The formula for kinetic energy is KE = 1/2 mv^2, where m is the mass of the object and v is its velocity.

Since the ball is initially at rest, the initial kinetic energy (KE_initial) is zero. So, the total mechanical energy (E_total) is equal to the initial potential energy (PE_initial) of the ball.

E_total = PE_initial
2.94 J = 1/2 (0.12 kg) * v^2

Rearranging the equation and solving for v, we get:

v^2 = (2 * 2.94 J) / (0.12 kg)
v^2 = 49 J/kg

Taking the square root of both sides yields:

v = sqrt(49 J/kg)
v = 7 m/s

Therefore, the velocity of the ball before it reaches the ground is 7 m/s.