Answer:[tex]5 m/s^{2}[/tex]
Explanation:
This problem is related to vertical motion, and the equation that models it is:
[tex]y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2}[/tex] (1)
Where:
[tex]y=0m[/tex] is the rock's final height
[tex]y_{o}=100 m[/tex] is the rock's initial height
[tex]V_{o}=15 m/s[/tex] is the rock's initial velocity
[tex]\theta=90\°[/tex] is the angle at which the rock was thrown (directly upwards)
[tex]t=10 s[/tex] is the time
[tex]g[/tex] is the acceleration due gravity in Planet X
Isolating [tex]g[/tex] and taking into account [tex]sin(90\°)=1[/tex] :
[tex]g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t)[/tex] (2)
[tex]g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s))[/tex] (3)
[tex]g=5 m/s^{2}[/tex] (4) This is the acceleration due gravity in Planet X
