a 65kg diver jumps off of a 10m tower. find the diver´s velocity when he hits the water?

Set potential energy equal to potential energy. i.e. mgh = (1/2)mv^2

where m=mass, g=gravity or 9.8m/s^2, h=height, v=velocity

In the beginning, the diver has potential energy (since the person is raised at a certain distance above ground level). As the diver falls, this distance decreases and the velocity picks up - thus kinetic energy accumulates.

Although one form of energy is transformed in another form of energy, energy is conserved throughout the process.

To find the diver's velocity when he hits the water, we can use the principle of conservation of energy. The initial potential energy of the diver at the top of the tower will be converted into kinetic energy just before hitting the water.

The potential energy (PE) of an object is given by the formula PE = mgh, where m is the mass (65 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the tower (10 m).

PE = mgh
PE = 65 kg * 9.8 m/s^2 * 10 m
PE = 6370 J

This potential energy will be converted entirely into kinetic energy (KE) just before hitting the water. The formula for kinetic energy is KE = 1/2 mv^2, where v is the velocity.

KE = PE
1/2 mv^2 = 6370 J

To find v, rearrange the equation:

v^2 = (2 * (PE/m))
v^2 = (2 * 6370 J) / 65 kg
v^2 = 195.69 m^2/s^2

Taking the square root of both sides, we find:

v = √(195.69 m^2/s^2)
v ≈ 14 m/s

Therefore, the diver's velocity when he hits the water will be approximately 14 m/s.