Explanation:
This is a situation related to vertical motion with constant acceleration, where the equation that will be usefull is:
[tex]y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}[/tex] (1)
Where:
[tex]y=0[/tex] is the final height of the ball (when it reaches the ground)
[tex]y_{o}=0[/tex] is the initial height of the ball (is also zero because is thrown from ground)
[tex]V_{o}[/tex] is the initial velocity of the ball
[tex]t=2.5 s[/tex] is the time the ball is in air (on Earth)
[tex]g_{E}=9.8 m/s^{2}[/tex] is the acceleration due to gravity on Earth
[tex]g_{M}=1.62 m/s^{2}[/tex] is the acceleration due to gravity on the Moon
Having this clear, let's solve (1) for the Earth:
[tex]0=0+V_{o}t-\frac{1}{2}gt^{2}[/tex] (2)
[tex]0=t(V_{o}-\frac{1}{2}gt^{2})[/tex] (3)
[tex]V_{o}=\frac{1}{2}gt^{2})[/tex] (4)
[tex]V_{o}=\frac{1}{2}(9.8 m/s^{2})(2.5 s)^{2})[/tex] (5)
[tex]V_{o}=12.25 m/s[/tex] (6) This is the initial velocity
Using this same velocity and equation (4) for the Moon:
[tex]12.25 m/s=\frac{1}{2}(1.62 m/s^{2})t^{2})[/tex] (7)
Finally finding [tex]t[/tex]:
[tex]t=15.12 s \approx 15 s[/tex]
