Question

AEROSPACE 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

634.675 ft

Step-by-step explanation:

After 1 second, distance = 2.65 ft

It falls 5.3 ft more than it did in the 1st second.

After 2 seconds, distance = 2.65 ft + 5.3 ft = 7.95 ft

It falls 5.3 ft more than it did in the 2nd second.

In the 2nd second, it fell by 5.3 ft. In the 3rd second, it falls 5.3 ft + 5.3 ft = 10.6 ft

After 3 seconds, distance = 7.95 ft + 10.6 ft = 18.55 ft

It follows that in the n-th second, it falls by 5.3(n - 1) ft.

In the 10th second, it falls by 5.3(10 - 1) = 5.3 × 9 = 47.7 ft.

Total distance fallen = 2.65 + [2.65 + (5.3 × 1)] + + [2.65 + (5.3 × 2)] + ... + [2.65 + (5.3 × 9)] = 2.65[1 + (5.3 × 1) + (5.3 × 2) + ... + (5.3 × 9)]

= 2.65[1 + 5.3(1 + 2 + ... + 9)] = 2.65[1 + (5.3 × 45)]

= 2.65(1 + 238.5) = 2.65 × 239.5 = 634.675 ft

Answer 2

d = 265 ft

Therefore, an object fall 265 ft in the first ten seconds after being dropped

Step-by-step explanation:

This scenario can be represented by an arithmetic progression AP.

nth term = a + nd

Where a is the first term given as 2.63 ft.

d is the common difference and is given as 5.3ft.

n is the particular second/time.

To calculate how far the object would fall in the first 10 seconds, we can derive it using the sum of an AP.

d = nth sum = (n/2)(2a+(n-1)d)

Where n = 10 seconds

a = 2.65 ft

d = 5.3 ft

Substituting the values we have;

d = (10/2)(2×2.65 + (10-1)5.3)

d = 265 ft

Therefore, an object fall 265 ft in the first ten seconds after being dropped