Can anyone please help...urgent.?

Question

A ball dropped from a height of 15m rebounds from the ground to a height of 10m. With each successive rebound, it rises two-thirds of the height of the previous rebounds. What total distance will it have travelled when it strikes the ground for the tenth time?

Answer 1

well, let's start counting:

1) for the first strike we know that that ball traveled 15m

1a) now the ball went up 10m (also given)

2) for the second strike it has to go down 10m

2.1) now this is where we start making calculations. we are given that the ball rises two thirds of the height of the previous rebound. so this time the ball rises 10*2/3 = 20/3 meters.

3) for the third strike the ball goes down 20/3m

3.1) now it rises to 20/3 * 2/3 = 40/9m

and so on...

you can see that the ball does the following route:

1. go down 15m

2. go up 10m and down 10m

3. go up 20/3m and down 20/3m

4. go up 40/9m and down 40/9m

.

.

.

10. go up 10*(2/3)^8 meters and down 10*(2/3)^8 meters

All this sums up to 481835/6561 which is about 73.44

Answer 2

Let x be the original 15m distance.

Let n be the number of bounces.

1st bounce distance = (2/3)^1 (15m) = 10m

2nd bounce distance = 2(2/3)^2 (15m) = 2x6.667m

3rd bounce distance = 2(2/3)^3 (15m)

4th bounce distance = 2(2/3)^4 (15m)

When you add all the bounces, it becomes the following

(15m)[(2/3)^1 + 2(2/3)^2 + 2(2/3)^3 + 2(2/3)^4 + 2(2/3)^5 + 2(2/3)^6 + 2(2/3)^7 + 2(2/3)^8 + 2(2/3)^9 + 2(2/3)^10)] = final distance.. whatever it is..

Hope this helps.

Answer 3

0.3901844231 metres

Answer 4

   this problem sames very difficult but it is really not.choosing to pile up data and sum them will really give wrong approximations  of the distance  and even boring.

        

1) the ball is dropped from height of 15m and it rebounds to height of 10m

2) it rebound height ratio=2/3 of previous heights.

     *note this can be solve as G.P with common ratio 2/3

     - number of terms=10 , first term a=15

      Here we look for sum of 10 of the G.P.this is the easiest way

      *but know here that the distance decreases in consecutively.

   15,(15x2)/3,(10x2)/3.......10 term

          Sn=a(1-r^n)/(1-r) for r<1

              =15(1-(2/3)^10)/(1-2/3)

              =15(1-(1024)/59049)x3

              =45(1-0.17341)

              =45(0.98265)

              =44.2196m  

    * the ball has to go up and come down which mean if it covers Xm up then it should cover 2Xm for up and down distance therefore multiply the above distances by 2 and minus 15m since it was dropped from there

             =>S= (44.2196X2)m

                 S=88.44m-15m

                  S=73.4385

       this is the approximate distance it has to cover from the 15m height to the tenth bounds.

       Alternatively one could multiply a=15 by 2 then minus 15m and working with in this we could had

                      S=   90(0.98265)-15

                      S=73.4385m APPROXIMATELY.

you can try it you self by considering this question as a G.P  with common ratio=2/3 and first term,a=(15x2) and n=10 and use the formula of the sum of a G.P with r<1 then minus the 15m and get the result as the distance it covers