A pole vaulter has 3500J of energy when he crosses a bar 7 metres high. How much kinetic energy will he have before he reaches the ground?

3500J

To find the kinetic energy of the pole vaulter before he reaches the ground, we need to calculate the potential energy he has at the top of the bar, and then subtract it from the total energy he has when crossing the bar.

The potential energy (PE) at a certain height is given by the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth), and h is the height.

Since we are given the potential energy at the height of 7 meters, we can rearrange the formula and solve for m:

PE = mgh
3500 J = m * 9.8 m/s^2 * 7 m

Now, we can solve for the mass (m):

m = 3500 J / (9.8 m/s^2 * 7 m)
m ≈ 55.8 kg

Now that we know the mass of the pole vaulter, we can calculate the kinetic energy (KE) he will have before reaching the ground. The formula for kinetic energy is KE = 1/2 * mv^2, where v is the velocity of the object.

To find the velocity, we need to use the concept of conservation of energy. The total energy at any point in time remains constant, so the total energy at the top of the bar is equal to the total energy before reaching the ground:

PE at the top = KE before reaching the ground

mgh = 1/2 * mv^2

Now, we can solve for the kinetic energy (KE) before reaching the ground:

KE = mgh - PE at the top
KE = mgh - 3500 J

Let's plug in the values we know:

KE = 55.8 kg * 9.8 m/s^2 * 7 m - 3500 J
KE ≈ 3857 J

Therefore, the pole vaulter will have approximately 3857 J of kinetic energy before reaching the ground.

0J kinetic energy