Spy Problem (For Calculus Class) Part 2?

Question

Our friend the spy (who escaped from the evil headquarters) is on a secret mission in space. Unfortunately, an encounter with an enemy agent has left him with a concussion which causes him to forget where he is. Fortunately, he remembers the formula for the height of a projectile h(t)=.5gt^2+vt+h where v and h are initial velocity and height. He does the following experiment to detirmine where he is.
1. 3 sec rock is rising reaches 108 ft
2. rock reaches max height of 121.5ft
What is value of g for this locale?
Mars:12 Venus:28 Earth 32 Moon 5.5 etc.

Answer 1

he landed on Pluto, which has been demoted to dwarf planet (is that politically correct? should it be little people planet, or Pluto-ette) the rock acted as a boomerang of sorts (what goes up must come down) he lofted it in a straight line, so it came straight down, (no trajectory figure given) knocking him out before he could finish the calculation using his solar powered ti-83

Answer 2

Suppose a rock is dropped from the same height, 121.5feet at that location. By symmetry the last 108 feet will take three seconds. Then the height equation becomes:
h = (1/2)gt^2 + v(zero)t + s(zero)
h = (1/2)gt^2 + 121.5
At the instant the rock reaches the ground:
0 = (1/2)gt^2 + 121.5
(1/2)gt^2 = -121.5
gt^2 = -243
g = -243/t^2
at 108 feet:
108 = (1/2)gt^2 + 121.5
(1/2)gt^2 = 108 - 121.5
(1/2)gt^2 = - 13.5
gt^2 = -27
g = -27/t^2
Let T = the time when the rock reaches 108 feet. Then the time when the rock reaches the ground = T + 3.
Set the two expressions for g equal to each other:
-27/T^2 = -243/(T + 3)^2
-27(T + 3)^2 = -243T^2
T^2 + 6T + 9 = 9T^2
8T^2 -6T -9 = 0
(2T - 3)(4T + 3) = 0
T = 3/2, T = -3/4 [extraneous]
g = -27/1.5^2 = -12feet/s^2 [He's on Mars]