contestada

A bouncy ball is dropped such that the height of its first bounce is 6.5 feet and each successive bounce is 78% of the previous bounce's height. What would be the height of the 10th bounce of the ball? Round to the nearest tenth (if necessary).

Respuesta :

putrishafaniscayah

Answer:

blablabla albalbalamakakms

jacob193

Answer:

Approximately [tex]0.7\; {\rm ft}[/tex].

Step-by-step explanation:

If the current bounce is of height [tex]h\; {\rm ft}[/tex], the next bounce would be of height [tex](78\%\, h)\; {\rm ft}[/tex], which is equal to [tex](0.78\, h)\; {\rm ft}[/tex].

It is given that the first bounce is of height [tex]6.5\; {\rm ft}[/tex]. Relative to this first bounce:

  • The [tex]n = 2[/tex] bounce is dampened [tex](2 - 1) = 1[/tex] time. The height of this bounce would be [tex]((0.78)\, 6.5)\; {\rm ft} = ((0.78)^{2 - 1} \, 6.5)\; {\rm ft}[/tex].
  • The [tex]n = 3[/tex] bounce is dampened [tex](3 - 1) = 2[/tex] times. The height of this bounce would be [tex](0.78)\, ((0.78)\, 6.5)\; {\rm ft} = ((0.78)^{3 - 1} \, 6.5)\: {\rm ft}[/tex].

In general, the [tex]n[/tex]th bounce would have been dampened [tex](n - 1)[/tex] times. The height of that bounce would be [tex]((0.78)^{n - 1}\, 6.5)\; {\rm ft}[/tex].

Thus, the [tex]10[/tex]th bounce would have been dampened [tex](10 - 1) = 9[/tex] times. The height of that bounce would be [tex]((0.78)^{10 - 1} \, 6.5)\; {\rm ft} \approx 0.7\; {\rm ft}[/tex] (rounded to the nearest tenth, one digit after the decimal point.)

Otras preguntas