Recent content by bolas

I need help with this one: \frac{1}{\cos(x)-1}dx I know there must be a way to rewrite it so that i can use 1 - cos^2 = sin^2. I've tried to multiply by cosx + 1 / cosx + 1 and that gets me further but the answer doesn't seem to be the same maybe because of simplification? Any hints...

For this problem based on the given information that th acceleration is 0.25g down the incline plane, can I assume then that the angle is... 0.25g = gsin(theta) theta = 14.47 deg ?

Gave up? :(

This is what I have so far.. Using equation of motion.. Vf^2 = Vi^2 + 2*a*x a = 0.25g = 0.25*(-9.8 m/s^2) = -2.45 m/s^2 Vf = 0 m/s Vi = ? x = s = ? 0 = Vi^2 + 2(-2.45 m/s^2)x 0 = Vi^2 - 4.9 m/s^2 * x -Vi^2 = -4.9 m/s^2 * x Vi^2 = 4.9 m/s^2 *x x = vi^2/4.9 m/s^2

"a ball is rolled up an incline (at an angle x with the horizontal). For the angle x and the ball involved, the acceleration of the ball is 0.25g and directed down the incline. If the ball is released with speed v, determine the distance it moved up the incline before reversing its direction as...