Velocity/speed question?

Question

There is a 250m difference in the altitude between the start and end of a ski-slope. A 64kg skier races down the slope. What is the speed of the skier at the bottom of the slope?

Could you please show working/steps? The answer is 31.3m/s, im just not sure how to do it. :) Thanks.

Answer 1

V=(gt) + V0, H=(1/2)9.8t^2,then 250=4.9t^2,t^2=51.02, t=7.143 sec.

V=9.8 X 7.143=70 mrter/sec.( It is independaent of his/her mass.)

Answer 2

V² = U² + 2*a*s

V² = 0 +2*9.81*250

V² = 4905

V = √4905

V = 70.m/s

Answer 3

I get the same -- 70 m/s

½mv² = mgh

v = √2gh

v = √2 * 9.81 * 250

70.0 m/s

Are you sure nothing is missing, such as the slope, and/or friction?

Answer 4

i don't know.

Answer 5

Times it 250 x 64=160 or 250/64/

Answer 6

They are booking.

Answer 7

V=?,s=250m,m=64Kg,g=10   V^2=U^2+2gs  V^2=0^2+2*10*250

V^2=5000  V=srqt5000  V=70.71m/s  this was what I got.

Answer 8

I don't think that answer is right

Conservation of Energy

Einit = m g h

Efinal = 1/2 m v^2

Einit = Efinal

1/2 m v^2 = m g h

v = Sqrt(2 g h)

maybe I'm missing something...

Actually either the book has a typo or you do because if the height is only 50 meters then the answer you gave comes out.

EDIT:  ROFL!  Yeah so text books have misprints ALL THE TIME.  Go with what you know, not what the book is telling you the answer is.