What is her speed when she gets to the bottom of the slope?

Her potential energy at the top of the slope = (mass) x (gravity) x (height)

Her kinetic energy at the bottom of the slope = (1/2) x (mass) x (speed) ².

The only possible way to handle this problem without making it

grossly complicated is to assume that the slope is frictionless.

Then ALL of the potential energy she had at the top shows up

as kinetic energy at the bottom, so we can write:

(1/2) x (mass) x (speed) ² = (mass) x (gravity) x (height)

Divide each side

by (mass) : (1/2) x (speed) ² = (gravity) x (height)

Isn't that interesting!

The speed at the bottom only depends on the height of the hill

and what planet she's skiing on. Her mass/weight doesn't matter.

A beanpole and a tons-o'-fun will both have the same speed when

they reach the bottom!

(1/2) x (speed) ² = (gravity) x (height)

Multiply each side

by 2 : speed² = (2) x (gravity) x (height)

Square root

each side: Speed = √ (2 g h)

= √ (19.6 h)

= about 4.43 x √ (height of the hill, meters)

Her speed to the bottom of the slope decreases to zero, since it seems she has halted.