How Does Gravity on Planet X Compare to Earth's?

[J89]

Homework Statement

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?

Homework Equations

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

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 :(

 

[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.

 

[nlightner]

I would approach this problem by first identifying the relevant equations and variables. In this case, the equations for projectile motion in the x and y directions are relevant, as well as the acceleration due to gravity (g). The variables that we know are the distances D and 2.76D, and the initial speed (V0) is the same on both Earth and Planet X.

Next, I would use the given information to set up a system of equations. Since we know that the initial speed and angle of launch are the same on both Earth and Planet X, we can set the equations for the y-direction equal to each other:

(V0sina0)t - 1/2gt^2 = (V0sina0)t - 1/2gt^2

From this, we can see that the only difference between the two scenarios is the value of g. So, we can set up a proportion:

D/g on Earth = 2.76D/g on Planet X

Solving for g on Planet X, we get:

g = (2.76D/g)(D/2.76D) = 2.76g

Therefore, the acceleration due to gravity on Planet X is 2.76 times the value of g on Earth. This makes sense, as the planet could have a higher mass or different composition, leading to a stronger gravitational pull.