A stationary roller coaster at the top of its path is 30m about the ground. The car and its passengers have a mass of 400kg. What is the velocity of the car when it reaches the ground lever? The formula for velocity is (Distance/time)

the gain in Kinetic energy is equal to the loss of potential energy

1/2 m v^2=mgh
solve for v

mass cancels out, irrelevant

m g h = (1/2) m v^2

2 g h = v^2

v = sqrt (2 g h)

v = sqrt (2*9.8*30)

v = 24.2 m/s

To calculate the velocity of the roller coaster at the ground level, you can use the principle of conservation of energy. This principle states that the total mechanical energy of a system remains constant unless acted upon by external forces.

Initially, when the roller coaster is at the top of its path, it has potential energy due to its height above the ground. At ground level, this potential energy is converted entirely into kinetic energy, given that there is no external force acting on the system.

So, the potential energy at the top of the path can be calculated using the formula:

Potential Energy = mass * gravitational acceleration * height

Potential Energy = 400 kg * 9.8 m/s^2 * 30 m = 117,600 Joules

Since this potential energy will be converted into kinetic energy at the ground level, we can equate the two:

Kinetic Energy = 1/2 * mass * velocity^2

Substituting the known values:

117,600 J = 1/2 * 400 kg * velocity^2

117,600 J = 200 kg * velocity^2

Now, solve for velocity:

velocity^2 = 117,600 J / 200 kg
velocity^2 = 588 m^2/s^2

Taking the square root of both sides, we get:

velocity = √(588 m^2/s^2)
velocity ≈ 24.25 m/s

Therefore, the velocity of the car when it reaches the ground level is approximately 24.25 m/s.