Answer:
d = 265 ft
Therefore, an object fall 265 ft in the first ten seconds after being dropped
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
Answer:
634.675 ft
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
