Recent content by WannaLearn

----------------------------------------------------------------------------------------- Thank you pastro. Step 1. 2.36g=G/r^2 ? I found a solution to this problem in one of the previous posts, but there are few things I do not understand. Solution: radius of orbit of Europa, rm =...

1. The problem: The moon Europa, of the planet Jupiter, has an orbital period of 3.55 days and an average distance from the center of the planet equal to 671,000 km. If the magnitude of the gravitational acceleration at the surface of Jupiter is 2.36 times greater than that on the surface of...

Thank you rock.freak. I appreciate your help. I am still stuck at part two. You referred to the kinematick equation x=x0+v0t+1/2at^2, is it to calculate the acceleration of the ball in motion? if yes, when I plug in the values for time and initial velocity, I get a -2.43 for acceleration. And...

Sorry for the confusion. The values mentioned at the beginning of the post were from the first time I performed the curve fit, and in the image attached I performed curve fit for the second time. But even if I take the values from the image A:-4.764, B:10.51, C:-3.916, when I divide -B/2A, I get...

http://www.ojancevic.com/images/parabola.jpg Thank you rock.freak, 2). The kinematic formula is x=x0+v0t+1/2at^2 3). I know that b/a would give me meters, and I assume by dividing -B/2A, I would get the highest point (the highest position the ball reaches). But when I look at the graph...

Homework Statement In the experiment that I have performed, the ball was thrown upward and caught once it reached its initial position. The data was recorded using the graphical analysis program. The graph produced is a parabola. The parabolic curve was generated showing the values of three...