FIND THE TOTAL DISTANCE IT HAS TRAVELLED WHEN IT HITS THE GROUND FOR THE 10TH TIME

## A BOUNCING TENISS BALL REBOUNCE EACH TIME TO A HEIGHT EQUAL TO HALF OF THE HEIGHT OF THE PREVIOUS. IF IT IS DROPPED FROM THE DISTANCE OF A 16 METRE

# 3 Answers

The total distance

= 16 + 16(1/2)*2 + 16(1/2^2)*2 + ... + 16(1/2^8)*2

= 16 + 16[1 + 1/2 + (1/2)^2 + ... + (1/2)^8]

= 16[ 1 + (1 - (1/2)^9)/(1-(1/2))]

= 16[3 - (1/2)^8] meters <==Answer {edited}

Attn:

1) I assume the ball is dropped vertically.

2) You need to multiply each of the last 8 terms by 2 because the ball goes up and down with the same distance in each bouncing.

16 + 8 + 8 + 4 + 4 + 2 + 2 + 1 + 1 + 1/2 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/16 + 1/16 + 1/32 + 1/32

= 16 + 16 + 8 + 4 + 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16

= 47 + 15/16

= 767/16 metres

Tennis ball touch down 10 times, and bounce 9 times.

Down ↓ * 16*,

*8, 4, 2, ...*Up ↑

**8, 4, 2, ...**So, we have two geometric sequences, were each number, after the first is twice less than previous. The only difference between them is

*.*

**16**Let's find the sum of the numbers of second geometric sequence ( Up )

**~~~~~~~~~~~~~~~~~**1 - r

^{n}

S

_{9}= a

_{1}• ————

1 - r

a

_{1}is the first term of geometric sequence,

r is the common ratio

n is the number of terms

**~~~~~~~~~~~~~~~~~~~~**a

_{1}= 8

r = 1/2

n = 9

1 - (1/2)

^{9}

S

_{9}= 8 · —————— =

1 - (1/2)

8 · [ 1 - (1/512) ] ÷ (1/2)

16 · (511 / 512) =

*511 / 32**is twice of "distance Up" and 16*

**The total distance**16 + 2 · (511/32) =

16 + (511/16) =

(256/16) + (511/16) =

**767/16 = 47.9375 m**