A) 2.64 H
The maximum height that the expelled rock can reach can be found by using the equation:
![v^2-u^2 = 2gd](https://tex.z-dn.net/?f=v%5E2-u%5E2%20%3D%202gd)
where
v = 0 is the velocity at the maximum height
u is the initial velocity
g is the acceleration of gravity
d is the maximum height
Solving for d,
![d=\frac{-u^2}{2g}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B-u%5E2%7D%7B2g%7D)
We see that the maximum heigth is inversely proportional to g. On the Earth,
and ![g=g_e = -9.81 m/s^2](https://tex.z-dn.net/?f=g%3Dg_e%20%3D%20-9.81%20m%2Fs%5E2)
So we can write:
![\frac{H}{H'}=\frac{g_m}{g_e}](https://tex.z-dn.net/?f=%5Cfrac%7BH%7D%7BH%27%7D%3D%5Cfrac%7Bg_m%7D%7Bg_e%7D)
where H' is the maximum height reached on Mars, and
is the acceleration of gravity on Mars. Solving for H',
![H' = \frac{g_e}{g_m}H = \frac{-9.81}{3.71}H=2.64 H](https://tex.z-dn.net/?f=H%27%20%3D%20%5Cfrac%7Bg_e%7D%7Bg_m%7DH%20%3D%20%5Cfrac%7B-9.81%7D%7B3.71%7DH%3D2.64%20H)
B) 2.64T
The time after which the rock reaches the maximum height can be found by using
![v=u+gt](https://tex.z-dn.net/?f=v%3Du%2Bgt)
where
v = 0 is the velocity at the maximum height
u is the initial velocity
Solving for t,
![t=\frac{v-u}{g}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bv-u%7D%7Bg%7D)
The total time of the motion is twice this value, so:
![t=2\frac{v-u}{g}](https://tex.z-dn.net/?f=t%3D2%5Cfrac%7Bv-u%7D%7Bg%7D)
So we see that it is inversely proportional to g.
On the Earth, t = T. So we can write:
![\frac{T}{T'}=\frac{g_m}{g_E}](https://tex.z-dn.net/?f=%5Cfrac%7BT%7D%7BT%27%7D%3D%5Cfrac%7Bg_m%7D%7Bg_E%7D)
where T' is the total time of the motion on Mars. Solving for T',
![T' = \frac{g_e}{g_m}T=\frac{-9.81}{-3.71}T=2.64T](https://tex.z-dn.net/?f=T%27%20%3D%20%5Cfrac%7Bg_e%7D%7Bg_m%7DT%3D%5Cfrac%7B-9.81%7D%7B-3.71%7DT%3D2.64T)