a skier, mass 61 kg is at the top of a hill and starts down with an initial speed of 5m/s. he does no work on the way down, assume that the hill surface is frictionless. find his speed at the bottom of the hill which is 10m below the top.

PE at top: m g h

Ke at the bottom equals PE at top.
1/2 m v^2=mgh
v=sqrt(2gh)

so 14m/s?

To find the skier's speed at the bottom of the hill, we can use the law of conservation of energy. According to this law, the total mechanical energy of a system remains constant if no external forces are acting on it.

In this case, since the surface of the hill is frictionless, no external forces are acting on the skier during the descent. Therefore, we can assume that the mechanical energy is conserved.

The mechanical energy of an object can be divided into two forms: potential energy (PE) and kinetic energy (KE). The formula for potential energy is given by:

PE = m * g * h

where m is the mass of the skier, g is the gravitational acceleration (9.8 m/s^2), and h is the height.

The formula for kinetic energy is given by:

KE = (1/2) * m * v^2

where m is the mass of the skier and v is the velocity.

At the top of the hill, the skier has only potential energy, and at the bottom of the hill, the skier has only kinetic energy.

Since energy is conserved, we can equate the initial potential energy to the final kinetic energy:

m * g * h(top) = (1/2) * m * v(bottom)^2

Now, we can solve for v(bottom):

v(bottom) = sqrt((2 * g * h(top))

Substituting the given values, we have:

v(bottom) = sqrt((2 * 9.8 m/s^2 * 10 m))

Simplifying this equation:

v(bottom) = sqrt(196)

v(bottom) = 14 m/s

Therefore, the skier's speed at the bottom of the hill is 14 m/s.