Question

If a rubber ball consistently bounces back ⅔ of the height from which it is dropped, what fraction of its original height will the ball bounce after being dropped and bounced four times without being stopped?
Kindly explain it too with solution

Answer 1

2

$( {2 \div 3})^{4}$

Answer 2

16/81 is the proper answer.

Step-by-step explanation:

• Each time the ball is dropped and it bounces back, it reaches  2/3 of the peak, it turned into dropped from.
• After the primary bounce, the ball reaches 2/3 of the peak from which it turned into dropped - allow us to name it the unique height. After the second bounce, the ball with attain 2/3 of the peak it might have reached after the primary bounce.
• So on the give-up of the second bounce, the ball might have reached  2/3× 2/3 of the original height = 4/9 the of the original height.
• After the 1/3 bounce, the ball attains 2/3 of the peak it might have reached after the second bounce.
• So, on the give-up of the 1/3 bounce, the ball might have reached  2/3× 2/3× 2/3= 8/27. After the fourth and closing bounce, the ball attains   2/3 of the peak it might have reached after the 1/3 bounce.

So, on the give-up of the closing bounce, the ball might have reached  2/3× 2/3×2/3× 2/3 of the original height = 16/81 of the original height.

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