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

Question

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

Answer 1

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.

Answer 2

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 - rⁿ
S₉  =  a₁ •  ————  
                   1 - r     

a₁ is the first term of geometric sequence,
r is the common ratio
n is the number of terms
~~~~~~~~~~~~~~~~~~~~
a₁ = 8
 r = 1/2
 n = 9
                   1 - (1/2)⁹     
S₉ =  8  ·  —————— =           
                   1 - (1/2)

        8 · [ 1 - (1/512) ] ÷ (1/2)
        16 · (511 / 512) = 511 / 32

The total distance is twice of "distance Up" and 16
16 + 2 · (511/32) =
16 + (511/16) =
(256/16) + (511/16) =
767/16 = 47.9375 m

Answer 3

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