An 60kg athlete wants to jump over a 2 m bar. How fast must she go the moment her foot leaves the ground in order to get her centre of mass up so high? I set KE of jumper to PE at top of bar to get 6.26m/s. Am I on the right track?

Yes, you are on the right track! To determine the necessary speed at which the athlete must leave the ground in order to clear the 2 m bar, you correctly equated the kinetic energy (KE) of the athlete to the potential energy (PE) at the top of the bar.

Let's break down the problem step by step:

1. Start by determining the potential energy at the top of the bar. The potential energy (PE) is given by the equation PE = m * g * h, where m is the mass of the athlete, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of the bar.

In this case, the height (h) is 2 m, and the mass (m) is given as 60 kg. So, the potential energy at the top of the bar is PE = 60 kg * 9.8 m/s² * 2 m = 1176 J.

2. Next, equate the potential energy (PE) to the kinetic energy (KE) at the moment the athlete's foot leaves the ground. The kinetic energy (KE) is given by the equation KE = (1/2) * m * v², where m is the mass of the athlete and v is the velocity.

Setting the potential energy equal to the kinetic energy, we get 1176 J = (1/2) * 60 kg * v².

3. Rearrange the equation to solve for the velocity (v). Multiply both sides by 2 and divide both sides by 60 kg to get v² = (2 * 1176 J) / (60 kg).

Simplifying, we have v² = 39.2 J/kg.

4. Finally, take the square root of both sides to solve for the velocity (v). Taking the square root of v² gives us v ≈ √39.2 m/s ≈ 6.26 m/s.

Therefore, to clear the 2 m bar, the athlete must leave the ground with a velocity of approximately 6.26 m/s.

In summary, you correctly used the principles of conservation of energy to equate the potential energy and kinetic energy. By rearranging the equation and solving for the velocity, you obtained the correct answer of approximately 6.26 m/s. Well done!