The ball would travel 3 times as high.
The equation for finding the displacement (vertical height in this case) of an object with a constant is:
##v^2 = u^2 -2as##
where ##s## is the displacement , ##u## is the initial velocity , ##v## is the final velocity , and ##a## is the acceleration .For more information about the resoning behind this formula, look here.
We can rearrange this formula to give:
##s = (v^2-u^2)/(2a)##
since we want to know about displacement (height).
When throwing a ball vertically, the velocity at the top of the throw (when the ball is highest) is 0 , so we can remove the ##v^2## part of the equation , as ##0^2 = 0##.
Now we can look at this equation in terms of Earth.
##s_(earth) = (-u^2)/(2a)##
The initial velocity ##u## remains constant in both cases, but the acceleration due to gravity on mars is one third the acceleration due to gravity on earth or ##1/3a##, so our equation for Mars is:
##s_(mars) = (-u^2)/(2xx 1/3 a)##
This can be rearranged to produce:
##s_(mars) = 3(-u^2)/(2a)##
From this we see that ##s_(mars) = 3(s_(earth))##.
Therefore the displacement or height (##s##) on mars is three times that on earth so the ball will travel 3 times as high .
