A skier starts from rest at the top of a 27 degree slope 1.5km long.

A skier starts from rest at the top of a 27 degree slope 1.5km long. Neglecting friction, how long does it take to reach the bottom?

okay first thing is to find out how high the skier is at the top of the hill in relation to when he finishes (bottom of the hill):

using sine, we get the equation: sin(27) = x/1500m
-> 1500(sin(27)) = x
-> x = 680.986m

Now we can say that the total mechanical energy will always be equal within the system because there is no friction.Therefore, the skier's potential energy at the top will equal the kinetic energy at the bottom.

Ep = mgh , Ek= 1/2(m)(v^2)

Since they will be the same when looking at the start and the end, mgh=1/2(m)(v^2) (we can cancel the masses out because there is an "m" on each side and they are equal.

We are left with gh=1/2(v^2)
->9.81(680.986)=1/2(x^2)
->6680.47=1/2(x^2)
-> x^2 = 13360.94
-> x= 115.59m/s

Now it is a basic kinematics equation. We are given:

acceleration: (n/a)
v1: 0m/s
v2: 115.59m/s
distance: 1500m
time: (?)

we arent given acceleration, so we use the equation d=1/2(v1+v2)t
manipulating the formula to solve for time, we get t= 2d/(v1+v2)

We then simply plug in the variables above:
-> t=2(1500)/(0+115.59)
-> t=3000/115.59
-> t=25.95s

I hope this helped!!! (and that it isnt wrong)