In 1992, Ukrainian Sergei Bubka used a short pole to jump to a height of 6.13 m. If the maximum potential

energy associated with Bubka was 4.80 kJ at the midpoint of his jump, what was his mass?

1 answer

To find the mass of Sergei Bubka, we can use the relationship between potential energy (PE) and mass (m) given by the equation:

\[ PE = mgh \]

Where:

  • \( PE \) is the potential energy (in joules)
  • \( m \) is the mass (in kilograms)
  • \( g \) is the acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2 \))
  • \( h \) is the height (in meters)

Given:

  • The maximum potential energy \( PE = 4.80 , \text{kJ} = 4800 , \text{J} \)
  • The height at the midpoint of the jump can be considered half the total jump height, so \( h = \frac{6.13 , \text{m}}{2} = 3.065 , \text{m} \)

We can rearrange the formula to solve for mass \( m \):

\[ m = \frac{PE}{gh} \]

Now, substituting in the known values:

\[ m = \frac{4800 , \text{J}}{9.81 , \text{m/s}^2 \times 3.065 , \text{m}} \]

Calculating the denominator:

\[ 9.81 \times 3.065 \approx 30.1 , \text{m}^2/\text{s}^2 \]

Now substituting this back in:

\[ m \approx \frac{4800}{30.1} \approx 159.73 , \text{kg} \]

Thus, Sergei Bubka's mass is approximately:

\[ \boxed{160 , \text{kg}} \]