A skier is accelerating down a theta = 33.0o hill at a =3.70m/s2?

Question

How long will it take her to reach the bottom of the hill, assuming she starts from rest and accelerates uniformly, if the elevation change is 339m?

I already found that the magnitude of her vertical acceleration is 2.02 m/s^2... I'm stuck and I'm desperate for help. 30 more min until the timer is out!

Answer 1

Use the distance formula relative to initial velocity and acceleration:
d = vt + 1/2at^2

You've already got the vertical acceleration, and can assume no initial velocity, so plug that in:
339 = 0 * t + 1/2 * 2.02 * t^2
339 = 2.02 t^2
t^2 = 339/2.02 = 167.8218
t = 13.0 sec.