Question

A ping pong ball has a 75% rebound ratio. When you drop it from a height of k feet, it bounces and bounces endlessly. If the height of the bounce follows geometric progression, find the general term that represent the situation in term of k. [2 marks] Hence, if the ball dropped from a height of 235ft, determine i. the highest height achieved by the ball after six bounces ii. the total distance travelled by the ball when it strikes the ground for 12th time [2 marks] [3 marks b) The number of cell double after each process of cell division every 2 hours. If there are 120 bacteria initially, how many bacteria will be there after one and half day? [4 marks]​

Answer 1

The movement of the ping pong ball follows a exponential progression.

• The expression for the k-th bounce is: $\mathbf{T_k = 235 \times (0.75)^k}$
• The height after the 6th bounce is 41.82 ft
• The total distance after the 12th bounce is 1584.84ft.
• There will be 31457280 bacteria after a day and a half.

1. Ping pong ball

The given parameters are:

$\mathbf{a = 235}$ --- the initial height

$\mathbf{r = 75\%}$ --- the common ratio

(a) The expression after bouncing k times

To do this, we make use of

$\mathbf{T_n = a \times r^n}$

So, we have:

$\mathbf{T_n = 235 \times (0.75)^n}$

In this case; k = n.

So, the equation becomes

$\mathbf{T_k = 235 \times (0.75)^k}$

Hence, the expression for the k-th bounce is: $\mathbf{T_k = 235 \times (0.75)^k}$

(b) The height after bouncing 6 times

This means that k = 6.

So, we have:

$\mathbf{T_k = 235 \times (0.75)^k}$

$\mathbf{T_6 = 235 \times (0.75)^6}$

$\mathbf{T_6 = 41.82}$

Hence, the height after the 6th bounce is 41.82 ft

(c) The total distance after the 12th bounce

First, we calculate the total height after the 12th bounce using:

$\mathbf{S_k = \frac{a(1 - r^k)}{1 - r}}$

So, we have:

$\mathbf{S_k = \frac{235 \times (1 - 0.75^{12})}{1 - 0.75}}$

$\mathbf{S_k = \frac{235 \times (0.968)}{0.25}}$

$\mathbf{S_k = 909.92}$

So, the total distance is:

$\mathbf{Distance = 2 \times S_k - a}$

$\mathbf{Distance = 2 \times 909.92 - 235}$

$\mathbf{Distance = 1584.84}$

Hence, the total distance after the 12th bounce is 1584.84ft.

(2) Cell division

The given parameters are:

$\mathbf{r = 2/2hr}$ --- rate

$\mathbf{a = 120}$ --- initial number of cells

$\mathbf{t = 18}$ --- number of 2 hours is a day and a half

So, the number of bacteria is:

$\mathbf{T_t = ar^t}$

This gives

$\mathbf{T_t = 120 \times 2^{18}}$

$\mathbf{T_t = 31457280}$

Hence, there will be 31457280 bacteria after a day and a half.

Read more about exponential functions at:

https://brainly.com/question/11487261