The height h(n) of a bouncing ball is an exponential function of the number n of bounces.

One ball is dropped and on the first bounce reaches a height of 6 feet. On the second

bounce it reaches a height of 4 feet.

PLEASE HURRY

Respuesta :

Answer:

The height of a bouncing ball is defined by [tex]h(n) = 6\cdot \left(\frac{4}{6} \right)^{n-1}[/tex].

Step-by-step explanation:

According to this statement, we need to derive the expression of the height of a bouncing ball, that is, a function of the number of bounces. The exponential expression of the bouncing ball is of the form:

[tex]h = h_{o}\cdot r^{n-1}[/tex], [tex]n \in \mathbb{N}[/tex], [tex]0 < r < 1[/tex] (1)

Where:

[tex]h_{o}[/tex] - Height reached by the ball on the first bounce, measured in feet.

[tex]r[/tex] - Decrease rate, no unit.

[tex]n[/tex] - Number of bounces, no unit.

[tex]h[/tex] - Height reached by the ball on the n-th bounce, measured in feet.

The decrease rate is the ratio between heights of two consecutive bounces, that is:

[tex]r = \frac{h_{1}}{h_{o}}[/tex] (2)

Where [tex]h_{1}[/tex] is the height reached by the ball on the second bounce, measured in feet.

If we know that [tex]h_{o} = 6\,ft[/tex] and [tex]h_{1} = 4\,ft[/tex], then the expression for the height of the bouncing ball is:

[tex]h(n) = 6\cdot \left(\frac{4}{6} \right)^{n-1}[/tex]

The height of a bouncing ball is defined by [tex]h(n) = 6\cdot \left(\frac{4}{6} \right)^{n-1}[/tex].

Answer:

9 (2/3) ² but insted of 2 its n

Step-by-step explanation: