On the Moon, a falling object falls just 2.65 feet in the first second after being dropped. Each second it falls 5.3 feet farther than in the previous second. How far would an object fall in the first ten seconds after being dropped?

Answer 1

You can either use a formula or construct a table. You should get the same answer.
If the falling object drops 2.65 feet the first second, the value of the accleration of gravity there is
g' = 5.3 ft/s^2.

In t = 10 seconds, use the formula
Y = (g'/2)*t^2, with
g' = 5.3 ft/s^2 and t = 10 s.

Y = 265 feet.

1st second: 2.65 ft
2nd second: 5.3 + 2.65 = 7.95 ft
3rd second: 7.95 + 5.3 = 13.25 ft
4th second: 18.55 ft
5th second: 23.85 ft
6th second: 29.15 ft
7th second: 34.45 ft
8th second: 39.75 ft
9th second: 45.05 ft
10th second: 50.35 ft
Total: 262.35 ft