# 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?

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:

${v}^{2} = {u}^{2} - 2 a s$

where $s$ is the displacement ,
$u$ is the initial velocity ,
$v$ is the final velocity ,
and $a$ is the acceleration .

We can rearrange this formula to give:

$s = \frac{{v}^{2} - {u}^{2}}{2 a}$

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.

${s}_{e a r t h} = \frac{- {u}^{2}}{2 a}$

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 $\frac{1}{3} a$, so our equation for Mars is:

${s}_{m a r s} = \frac{- {u}^{2}}{2 \times \frac{1}{3} a}$

This can be rearranged to produce:

${s}_{m a r s} = 3 \frac{- {u}^{2}}{2 a}$

From this we see that ${s}_{m a r s} = 3 \left({s}_{e a r t h}\right)$.

Therefore the displacement or height ($s$) on mars is three times that on earth so the ball will travel 3 times as high .