Question

A baseball has a mass of 0.145 kg and is 12 m above the ground. If it has a mechanical energy of 105.8 J, how fast is it moving?

17 m/s
35 m/s
89 m/s
130 m/s

Answer 1

To find the speed of the baseball when we know its mechanical energy, we can use the formula for mechanical energy, which is the sum of kinetic energy (KE) and potential energy (PE):

\[
E_{\text{mechanical}} = KE + PE
\]

The potential energy (PE) of the baseball at a height \( h \) is given by the formula:

\[
PE = mgh
\]

where:
- \( m \) is the mass of the baseball (0.145 kg),
- \( g \) is the acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height (12 m).

First, let's calculate the potential energy:

\[
PE = 0.145 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 12 \, \text{m}
\]

\[
PE = 0.145 \times 9.81 \times 12
\]

Calculating that step-by-step:

1. \( 0.145 \times 9.81 \approx 1.42045 \)
2. \( 1.42045 \times 12 \approx 17.0454 \)

So, the potential energy \( PE \) is approximately \( 17.05 \, \text{J} \).

Now, let's use the total mechanical energy to find the kinetic energy:

\[
E_{\text{mechanical}} = KE + PE \implies KE = E_{\text{mechanical}} - PE
\]

Substituting the values:

\[
KE = 105.8 \, \text{J} - 17.05 \, \text{J} \approx 88.75 \, \text{J}
\]

Now, the kinetic energy (KE) is given by the formula:

\[
KE = \frac{1}{2} mv^2
\]

We can rearrange this to solve for \( v \):

\[
v = \sqrt{\frac{2 \times KE}{m}}
\]

Substituting in the values:

\[
v = \sqrt{\frac{2 \times 88.75 \, \text{J}}{0.145 \, \text{kg}}}
\]

Calculating the right side step-by-step:

1. \( 2 \times 88.75 \approx 177.5 \)
2. \( \frac{177.5}{0.145} \approx 1224.14 \)
3. \( \sqrt{1224.14} \approx 34.96 \)

Thus, \( v \approx 35 \, \text{m/s} \).

The speed of the baseball is approximately \( \textbf{35 m/s} \).