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?
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
To determine how far an object would fall in the first ten seconds after being dropped on the Moon, we can use the formula for the displacement of an object under constant acceleration:
displacement = initial displacement + (initial velocity × time) + (0.5 × acceleration × time^2)
In this case, we are given the initial displacement as 2.65 feet and the acceleration as 5.3 feet per second squared. We need to calculate the displacement for each second from 1 to 10 and then sum them up to get the total distance.
Let's calculate the distance for each second:
For the first second (t = 1):
displacement = 2.65 + (0 × 1) + (0.5 × 5.3 × 1^2) = 2.65 + 0 + 2.65 = 5.3 feet
For the second second (t = 2):
displacement = 2.65 + (0 × 2) + (0.5 × 5.3 × 2^2) = 2.65 + 0 + 10.6 = 13.25 feet
For the third second (t = 3):
displacement = 2.65 + (0 × 3) + (0.5 × 5.3 × 3^2) = 2.65 + 0 + 28.35 = 30 feet
We continue this calculation for each second until the tenth second and add up all the distances:
For the fourth second (t = 4):
displacement = 2.65 + (0 × 4) + (0.5 × 5.3 × 4^2) = 2.65 + 0 + 42.4 = 45.05 feet
For the fifth second (t = 5):
displacement = 2.65 + (0 × 5) + (0.5 × 5.3 × 5^2) = 2.65 + 0 + 66.25 = 68.9 feet
For the sixth second (t = 6):
displacement = 2.65 + (0 × 6) + (0.5 × 5.3 × 6^2) = 2.65 + 0 + 94.8 = 97.45 feet
For the seventh second (t = 7):
displacement = 2.65 + (0 × 7) + (0.5 × 5.3 × 7^2) = 2.65 + 0 + 128.45 = 131.1 feet
For the eighth second (t = 8):
displacement = 2.65 + (0 × 8) + (0.5 × 5.3 × 8^2) = 2.65 + 0 + 168.8 = 171.45 feet
For the ninth second (t = 9):
displacement = 2.65 + (0 × 9) + (0.5 × 5.3 × 9^2) = 2.65 + 0 + 214.65 = 217.3 feet
For the tenth second (t = 10):
displacement = 2.65 + (0 × 10) + (0.5 × 5.3 × 10^2) = 2.65 + 0 + 266 = 268.65 feet
Now, we add up all the distances from each second to get the total distance:
Total distance = 5.3 + 13.25 + 30 + 45.05 + 68.9 + 97.45 + 131.1 + 171.45 + 217.3 + 268.65 = 1038.55 feet
Therefore, an object would fall approximately 1038.55 feet in the first ten seconds after being dropped on the Moon.