A) 2.64 H
The maximum height that the expelled rock can reach can be found by using the equation:
[tex]v^2-u^2 = 2gd[/tex]
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,
[tex]d=\frac{-u^2}{2g}[/tex]
We see that the maximum heigth is inversely proportional to g. On the Earth,
[tex]d=H[/tex] and [tex]g=g_e = -9.81 m/s^2[/tex]
So we can write:
[tex]\frac{H}{H'}=\frac{g_m}{g_e}[/tex]
where H' is the maximum height reached on Mars, and [tex]g_m = -3.71 m/s^2[/tex] is the acceleration of gravity on Mars. Solving for H',
[tex]H' = \frac{g_e}{g_m}H = \frac{-9.81}{3.71}H=2.64 H[/tex]
B) 2.64T
The time after which the rock reaches the maximum height can be found by using
[tex]v=u+gt[/tex]
where
v = 0 is the velocity at the maximum height
u is the initial velocity
Solving for t,
[tex]t=\frac{v-u}{g}[/tex]
The total time of the motion is twice this value, so:
[tex]t=2\frac{v-u}{g}[/tex]
So we see that it is inversely proportional to g.
On the Earth, t = T. So we can write:
[tex]\frac{T}{T'}=\frac{g_m}{g_E}[/tex]
where T' is the total time of the motion on Mars. Solving for T',
[tex]T' = \frac{g_e}{g_m}T=\frac{-9.81}{-3.71}T=2.64T[/tex]
