A ball rebounds 1/2 of the height from which it is dropped. It is dropped 56m from the building and it kept bouncing. How far would the ball have travelled up and down when it strikes the ground for the 6th time?
make a sketch of the first few bounces and you will see:
1 bounce --- 56 m
2nd bounce = 2(28) = 56
3rd bounce = 2(14) = 28
4th bounc = 2(7) = 14
5th bounce = 2(3.5) = 7
6th bounce = 2(1.75) = 3.5
add them up to get ..... 164.5
or:
from the 2nd on , we have a geometric sequence with
a = 56, r = 1/2
We need the sum of
1st term + 5 terms of the GS
= 56 + 56(1 - (1/2)^5)/(1 - 1/2)
= 56 + 56( (31/32) / (1/2) )
= 56 + 56(31/16)
= 56 + 217/2
= 164.5
Thank you
Gak
Well, well, well, look who's bouncing back for another question! Let's do some calculations, shall we?
If the ball rebounds half of the height each time it drops, then after the first bounce, it would reach a height of 56m/2 = 28m.
Now, for each subsequent bounce, the ball will reach a height of half the previous bounce. So, for the second bounce, it would reach a height of 28m/2 = 14m. Are you following along?
For the third bounce, it would reach a height of 14m/2 = 7m, and so on.
Now, let's calculate the total distance the ball has traveled for the 6th bounce. To do that, we need to add up the distances traveled both up and down.
For the upward journeys: 56m + 28m + 14m + 7m + 3.5m + 1.75m = 110.25m
For the downward journeys: 28m + 14m + 7m + 3.5m + 1.75m = 54.25m
So, when the ball strikes the ground for the 6th time, it would have traveled a total distance of 110.25m upward and 54.25m downward.
That's one bouncy ball!
To find the total distance the ball has traveled when it strikes the ground for the 6th time, we need to consider both the upward and downward distances it travels during each rebound.
First, let's calculate the maximum height the ball reaches after the first drop. The ball rebounds 1/2 of the height from which it is dropped, so it rebounds to a height of 56m x 1/2 = 28m.
The total distance traveled by the ball during the first bounce is the sum of the upward distance (28m) and the downward distance (56m) which is 28m + 56m = 84m.
For subsequent bounces, the ball will reach a maximum height of 28m x 1/2 = 14m after each bounce. However, since the ball strikes the ground for the 6th time, there are only 5 bounces. Therefore, we need to calculate the total distance traveled during these 5 bounces.
Considering the upward and downward distances during each bounce, the total distance traveled in one full bounce is 14m (upward) + 28m (downward) = 42m.
To find the total distance traveled during the 5 bounces, we multiply the distance traveled in one full bounce (42m) by the number of bounces (5): 42m x 5 bounces = 210m.
Adding the distance traveled during the first drop (84m), the total distance traveled when the ball strikes the ground for the 6th time is 210m + 84m = 294m.
Therefore, when the ball strikes the ground for the 6th time, it would have traveled a total distance of 294 meters (upward and downward combined).