Question

Kelsie Ahbe (72 kg ) was an Indiana Hoosier who competed in the 2016 Olympics in the pole vault competition. Her center of mass (COM) was 1.0 m above the ground as she ran toward the pole vault pit. For her best successful attempt, her final velocity the instant before she took off from the ground was 4.2 m/s. At the top of her jump, all of her mechanical energy was converted into potential energy.

(a) What was the height of Kelsie's winning jump, assuming the peak height of her COM while in the air was the same height as the bar?

Answer 1

The height of Kelsie's winning jump is approximately 0.907 meters.

Step-by-step explanation:

To calculate the height of Kelsie's winning jump, we can use the principle of conservation of mechanical energy. The mechanical energy at the top of her jump is equal to the gravitational potential energy when her center of mass is at its maximum height. We can equate the initial kinetic energy, which is given by the formula KE = 0.5 * m * v^2, to the gravitational potential energy, which is given by the formula PE = m * g * h. Rearranging the equation for gravitational potential energy, we get h = KE / (m * g).

Plugging in the values, we have KE = 0.5 * 72 kg * (4.2 m/s)^2 = 635.04 J. The gravitational acceleration, g, is approximately 9.8 m/s^2. Substituting these values into the formula for height, we get h = 635.04 J / (72 kg * 9.8 m/s^2) ≈ 0.907 m.

Therefore, the height of Kelsie's winning jump is approximately 0.907 meters.