Acceleration due to gravity on Mars is about one—third that on Earth. Suppose you throw a ball upward with the same velocity on Mars as on Earth. How would the balls maximum height compare to that on Earth?

1 Answer
Nov 10, 2015

The ball would travel 3 times as high.

Explanation:

The equation for finding the displacement (vertical height in this case) of an object with a constant acceleration 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 .