A ball on the end of a string is whirled around in a horizontal circle of radius .487 m. The plane of the circle is 1.06 m above the ground. The string breaks and the ball lands 2.26 m away from the point on the ground directly beneath the ball's location when the string breaks. The acceleration for gravity is 9.8 m/s^2. Find the centripetal acceleration of the ball during its circular motion. Answer in units of m/s^2. a[c] = (v^2)/r = (v^2)/.487 I need to find the velocity. I'm imagining a disk (representing the circular path of the ball before the string is broken) 1.06 m off the ground, and the ball landing on the ground 2.26 m away from the point on the ground beneath the edge of this disk. I'm not sure if my train of thought is correct however, for the following calculations: I assume the initial y component of velocity is zero because this disk is completely horizontal. However, once the string breaks, free-fall acceleration acts upon it. So, y = v[yi]t + .5a(t^2) => -1.06 = (0)t -4.9(t^2) t = (1.06/4.9)^(.5) This would be the time it took for the ball to get where it is. Since I am considering the horizontal disk, however, I don't need y in my calculations. But, I know that 2.26 is a horizontal distance, and, if divided by t, will give me the x component of velocity. I get about 4.86 m/s. Then, squaring the term and dividing by the radius (.487), my result is 48.48 m/s^2. I am uncomfortable about these assumptions I am making, however, so an explanation as to why these things can be assumed would be greatly appreciated. Thank you!