1. Feb 14, 2009

### J89

1. The problem statement, all variables and given/known data
Inside a starship at rest on earth, a ball rolls off the top of a horizontal table and lands at a distance D from the foot of the table. This starship lands now at a planet called Planet X. The commander Captain Cudos rolls the same ball off the same table with the same initial speed as on earth and finds that it lands a distance of 2.76D from the foot of the table. What is the acceleration due to gravity on Planet X?

2. Relevant equations
x=(V0cosa0)t
y=(V0sina0)t - 1/2gt^2
Vx=V0cosa0
Vy=V0sina0-gt

3. The attempt at a solution

I made up the initial speed since it was not given and assumed a time. and used y=(V0sina0)t - 1/2gt^2...came out completely wrong :(

2. Feb 14, 2009

### ll_lordomega

I made up a situation as well. I let the height of the table = 1.5 m and the horizontal velocity of the ball be 1 m/s. I then used the kinematic equation x= x(initial) + v(initial)t + .5at^2. I analyzed the vertical data for earth to see how long it was in the air, and got .55 sec. I then used the same kinematic for the x direction, except this time I used that time to see how far it went. I got .55 m. This is our "D" value. So, on planet x it travel 2.76 D. We get 1.52 m for the distance traveled for the ball. Since I choose a velocity horizontally of 1 m/s, that ball on planet x traveled for 1.52 sec before hitting the ground. Now, analyze the y direction using yet again the same kinematic and we can solve for a, or the acceleration due to gravity on planet x. Assuming i didn't mess up anything, it should be 1.3 m/s^2.