A ball bounces two-thirds of the distance it falls. If it is dropped from a height of 10 meters, how far does it move before hitting the floor for the fourth time?
hi Reiny,
U solved absolutely wrong, plz Ali follow this method
Solution:
For Downward Distance:-
a=10 (first term)
r=2/3 (common ratio)
n=4 (no. of terms)
s=? (sum of 4th terms)
Sn= a(1-r^n)/1-r (because r is less than1)
S= 10(1-(2/3)^4/1-(2/3)= 650/27
For Upward Distance
a=10*2/3=>20/3
r=2/3
n=3
S=?
S=a(1-r^n)/1-r
S=20/3(1-(2/3)^4/1-(2/3)
S=380/27
Now, add downward distance & upward distance for calculating total distance
650/27 + 380/27 = 1030/27
Distance traveled :
first bounce -- 10 m
2nd bounce -- (2/3)(10) or 20/3 both up and down
3rd bounce -- (2/3)(20/3) or 40/9 both up and down
4th bounce -- (2/3)(40/9) or 80/27 both up and down
total distance
= 10 + 2(20/3) + 2(40/9) + 2(80/27)
= 886/27 m
or appr 32.8 m
You are Right
Well, if the ball bounces two-thirds of the distance it falls, we can imagine the ball's journey as a series of falling and bouncing distances. Let's break it down.
First, the ball falls from a height of 10 meters, so it travels 10 meters downwards. It then bounces back up two-thirds of that distance, which is (2/3) * 10 = 6.67 meters.
Now, the ball is at a height of 6.67 meters. It falls this distance, travels 6.67 meters downwards, and then bounces back up two-thirds of that distance, which is (2/3) * 6.67 = 4.44 meters.
The ball keeps repeating this pattern of falling and bouncing. So, for the fourth time, it would move a cumulative distance of:
10 + 6.67 + 6.67 + 4.44 = 27.78 meters.
So, before hitting the floor for the fourth time, the ball would have moved approximately 27.78 meters. Just make sure to watch out for any clowns trying to juggle these bouncing balls!
To solve this problem, we need to determine the total distance the ball travels before hitting the floor for the fourth time.
Let's break down the motion of the ball.
When the ball is dropped from a height of 10 meters, it covers its full distance on the first bounce. Since the ball bounces two-thirds of the distance it falls, it reaches 10 * (2/3) = 20/3 meters on the first bounce.
On the second bounce, the ball falls from the height it reached on the first bounce. Again, it covers two-thirds of this distance. So, on the second bounce, it reaches (20/3) * (2/3) = (40/9) meters.
On the third bounce, the ball falls from the height it reached on the second bounce. Using the same pattern, it covers two-thirds of this distance. So, on the third bounce, it reaches (40/9) * (2/3) = (80/27) meters.
Finally, on the fourth bounce, the ball falls from the height it reached on the third bounce. It covers two-thirds of this distance. So, on the fourth bounce, it reaches (80/27) * (2/3) = (160/81) meters.
To find the total distance the ball moves before hitting the floor for the fourth time, we add up the distances covered on each bounce.
Total distance = 10 + (20/3) + (40/9) + (80/27) + (160/81)
Simplifying the fractions and adding, we get:
Total distance = 10 + 6.67 + 4.44 + 2.96 + 1.98
Total distance ≈ 25.05 meters
Therefore, the ball moves approximately 25.05 meters before hitting the floor for the fourth time.