Question

What is the translational speed of a bowling ball on a 0.76 m vertical rise?

Answer 1

To determine the translational speed of a bowling ball after rising vertically by 0.76 meters, we can use the principles of conservation of energy. Specifically, we can assume that the potential energy gained from the vertical rise is converted into kinetic energy.

The potential energy (PE) gained by the bowling ball as it rises can be calculated using the formula:

\[
PE = mgh
\]

where:
- \( m \) = mass of the bowling ball (in kg),
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) = height (0.76 m).

The kinetic energy (KE) of the bowling ball is given by:

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

where \( v \) is the translational speed.

According to the conservation of energy principle, the potential energy gained is equal to the kinetic energy lost (if we consider the start and end of the rise):

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

Here, we can see that the mass \( m \) cancels out from both sides of the equation:

\[
gh = \frac{1}{2}v^2
\]

To solve for \( v \), rearranging gives us:

\[
v^2 = 2gh
\]

Now substituting \( g = 9.81 \, \text{m/s}^2 \) and \( h = 0.76 \, \text{m} \):

\[
v^2 = 2 \cdot 9.81 \, \text{m/s}^2 \cdot 0.76 \, \text{m}
\]

Calculating this:

\[
v^2 = 2 \cdot 9.81 \cdot 0.76 \approx 14.9032
\]

Now taking the square root to find \( v \):

\[
v \approx \sqrt{14.9032} \approx 3.86 \, \text{m/s}
\]

Thus, the translational speed of the bowling ball after rising vertically by 0.76 m is approximately **3.86 m/s**.