Question

A 0.15-kg ball is dropped from the top of a 150-m building. What is the kinetic energy of the ball when it passes the sixteenth floor at a height of 63 m? (Ignore air resistance.)

Answer 1

The kinetic energy of the ball when it passes the sixteenth floor at a height of 63 m is approximately
`\(432 \, \text{J}\).`

Step-by-step explanation:

The kinetic energy
`(\(KE\))` of an object in free fall can be calculated using the equation:

`\[ KE = (1)/(2) m v^2 \]`

where \(m\) is the mass of the object and
`\(v\)` is its velocity. The velocity can be determined using the conservation of mechanical energy, which states that the sum of potential energy
`(\(PE\))` and kinetic energy remains constant in the absence of non-conservative forces like air resistance.

Initially, the ball has gravitational potential energy equal to its mass times the gravitational acceleration
`(\(mgh\))` , where
`\(m\)` is the mass,
`\(g\)` is the gravitational acceleration (approximately
`\(9.8 \, \text{m/s}^2\))` , and
`\(h\)` is the height. As the ball falls, this potential energy is converted to kinetic energy.

At the sixteenth floor
`(\(h = 63 \, \text{m}\))` , the potential energy is
`\(mgh\)`, and the kinetic energy is
`\((1)/(2)mv^2\)`. Equating the initial potential energy to the sum of kinetic and potential energy at the sixteenth floor, we can solve for the kinetic energy:

`\[ mgh = (1)/(2)mv^2 \]`

Solving for
`\(KE\)`, substituting the given values
`(\(m = 0.15 \, \text{kg}\)` and
`\(h = 63 \, \text{m}\)),`we find
`\(KE \approx 432 \, \text{J}\).` This result represents the kinetic energy of the ball when it passes the specified height.