Potential/Kinetic Energy Again.

[omgitsmonica]

Homework Statement

A skier, of mass 50 kg, pushes off the top of a hill with an initial speed of 5 m/s. Neglecting friction, how fast in m/s will he be moving after dropping 17 m in elevation? (Neglect friction, use g=10m/s2)

Homework Equations

GPE = MGH
KE =(1/2)mv^2

The Attempt at a Solution

KE = (1/2)mv^2
KE = (1/2)(50)(25)
KE = 625

GPE = mgh
625 = (50)(10)h
625 = 500h
1.25 = h

If he's only 1.25 meteres up, I can't drop him seventeen meters, so I know I'm doing something wrong. Please help.

Also. I'm sorry to post so often, but this homework is killing me.

 

[LowlyPion]

Homework Statement

A skier, of mass 50 kg, pushes off the top of a hill with an initial speed of 5 m/s. Neglecting friction, how fast in m/s will he be moving after dropping 17 m in elevation? (Neglect friction, use g=10m/s2)

Homework Equations

GPE = MGH
KE =(1/2)mv^2

The Attempt at a Solution

KE = (1/2)mv^2
KE = (1/2)(50)(25)
KE = 625

GPE = mgh
625 = (50)(10)h
625 = 500h
1.25 = h

If he's only 1.25 meteres up, I can't drop him seventeen meters, so I know I'm doing something wrong. Please help.

Also. I'm sorry to post so often, but this homework is killing me.

Unfortunately you've taken things a little backwards. The skier starts at 17m.

That gives you m*g*h of potential energy turned into KE at the bottom.
That yields 50*10*17 that goes into kinetic energy. 8500N-m
You also start out with KE = to 1/2m*v² = 625N-m

So at the bottom he has a Total of 9125N-m

That's all KE which again is 1/2m*v²
So... 9125 = 1/2m*v² = 1/2*(50)*v²
v² = 9125*2/50 = 365
v= 19.1m/s