physics question?

Question

A professional skier reaches a speed of 56 m/s on a 30 degree slope. ignoring friction, what was the minimum distance along the slope the skier would have had to travel, starting from rest?
The answer is 320 m, but I need to figure out how to get there.

Answer 1

There are several ways to solve this problem. I prefer the method which employs the conversion of gravitational potential energy (PE) into kinetic energy (KE) as the object falls a certain distance.

The change in KE comes at the expense of a change in PE.

KE = 1/2 mv^2
PE = mgh

The skier gains KE as he slides down the slope and looses PE.

The final KE of the skier is:
KE = 1/2 m (56 m/s)^2
We do not know the value for mass, but we will see that this does not matter.
Setting the change in KE equal to the change in PE,
PE = mgh = 1/2 m (56 m/s)^2
the masses cancel out and g is the gravitational acceleration constant (9.81 m/s^2).
We can solve for the vertical distance, h, the skier fell to be:
h = (56 m/s)^2 / (2 * (9.81 m/s^2))
h = 160 meters

This is the vertical distance remember, not the actual distance traveled.
Since the slope is at an angle of 30 degrees, we must divide the vertical distance by the sin(30) to get the overall distance traveled.

sine (theta) = opp / hyp
hyp = opp / sin (theta)
where hyp is the actual distance traveled, opp is the vertical distance, and theta is the angle

d = 160 / sin (30) = 160 / .5 = 160 * 2
d = 320 meters along the slope

Answer 2

loss in potential energy = gain in kinetic energy

suppose the skier with mass m ski down a height of H , then the loss in potential energy is mgH

its gain in kinetic E is = 1/2 * m * v^2

=> mgH = 1/2 m v^2

=> H = v^2/2g = 56^2/(2x9.81) = 160 m

if the distance travel is L then

L = H/sin 30 = 320 m (think bout right angle triangle and recall those trigo stuff)

Answer 3

Force on the skier is gravity...and on a 30 degree slope so the gravity is reduced to 9.8m/s^2(sin 30) = 4.9m/s^2

V(final) = (2as)^.5

s = 320

Answer 4

although weigh is not involved but here is how you find acceleration.
w=mg
Fw(x)= sin(30)mg

a=Fnet/mass
because Fw(x) is the net force we have
a= sin(30)mg / m
the mass cancel out we left with:
a=sin(30)g
a=4.9m/s^2

Vf^2=2ad+Vi^2
(56m/s)^2=2(4.9m/s^2)d
3136m^2/s^2= (9.8m/s^2)d
d=320m