Answer:
[tex]a_X=1.29m/s^2[/tex]
Explanation:
Horizontally we will use the equation [tex]t=\frac{d}{v}[/tex], where t is the time of fall of the ball, d is the distance from the foot of the table and [tex]v[/tex] the initial (horizontal only) speed of the ball.
Vertically we will use the equation [tex]h=\frac{at^2}{2}[/tex] (since the initial vertical speed is zero), where h is the height of the table and a the acceleration of gravity.
Combining both:
[tex]h=\frac{at^2}{2}=\frac{ad^2}{2v^2}[/tex]
Better written for our case as [tex]2hv^2=ad^2[/tex] since the variables on the left are the same for both the Earth and Planet X, so we know that what appears on the right hand side must be the same for both also.
This means that [tex]a_Ed_E^2=a_Xd_X^2[/tex], which means:
[tex]a_X=a_E(\frac{d_E}{d_X})^2[/tex]
For our values:
[tex]a_X=g(\frac{D}{2.76D})^2=\frac{g}{2.76^2}=\frac{9.8m/s^2}{2.76^2}=1.29m/s^2[/tex]
