Find "g" Value on Distant Planet: 6.501m/s^2

[climb515c]

Homework Statement

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"

Homework Equations

y=V(initial)(t) - 1/2(g)(t)^2

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.

 

[SHISHKABOB]

oh no, a difference of .004 is seriously no problem at all. The difference in your answer and the key's answer probably lies in how you rounded numbers in various parts of your process.

Really, don't worry about a difference that small. If you get an answer that is within 10% of the book's answer, you can assume that the difference is negligible.