Question

A ball is dropped and bounces to a height of 5 feet. The ball rebounds to 70% of its previous height after each bounce The function that could be used to model the nth term in the sequence of heights of the ball after the initial drop is

Answer 1

Answer:

$T_n = 7.143(0.7^n)$

Step-by-step explanation:

If a ball bounces to a height of 5feet, the initial drop = 5feet

If the ball rebounds to 70% of its previous height after each bounce, the next height after rebouncing will be expressed as;

= 70% of 5

= 70/100 * 5

= 0.7*5

= 3.5 feet

The next height will be 0.7*3.75 = 2.45 and so on

The  heights wil form a sequence as thus;

5, 3.5, 2.45...

This sequence forms a geometric progression.

To get the function that could be used to model the nth term in the sequence of heights of the ball after the initial drop, w will find the nth term of the sequence

$T_n = ar^{n-1}$

a is the first term of the sequence

n is the number of terms

r is the common ratio

From the sequence:

a = 5

$r = \frac{3.5}{5} = \frac{2.75}{3.5} = 0.7$

Substitute the given values into the formula

$T_n = 5(0.7)^{n-1}\\T_n = 5(\frac{0.7^n}{0.7^1} )\\T_n = \frac{5}{0.7}\times 0.7^n \\T_n = 7.143(0.7^n)$

The nth tern=m required is $T_n = 7.143(0.7^n)$