1. The problem statement, all variables and given/known data An astronaut on a distant planet drops a rock from the top of a cliff and observes it takes 2.532s to hit the base of the cliff. Then takes a 2nd rock and throws it straight up with a speed of 17.81m/s so that it reaches a height "h" above the cliff before falling to the base of the cliff, taking 6.470s to hit the ground. Air resistance is negligible, (one dimensional problem). Find the value of "g" 2. Relevant equations y=V(initial)(t) - 1/2(g)(t)^2 3. The attempt at a solution Being the 1st rock is just dropped and has zero initial velocity, I used y=-1/2(g)(t)^2, solving for "g" I got (g= y / -3.205512(s^2)). Then I plugged this value of "g" into the equation for the 2nd rock. y=(17.81m/s)(6.470s) - 1/2( y / -3.205512(s^2))(6.470s)^2 y=(115.2307m) + 6.529518529y -5.529518529y = 115.2307m y=-20.83919231m plugging this into the (g = y / -3.205512(s^2)) I find "g" = 6.501m/s^2 It feels like I missed something or skipped something important. The solution key has the answer as 6.505m/s^2.