A boy plays with a ball and he drops it from a height of 1.5 m. Every time the ball hits the ground, it bounces back to attain a height $\frac{4}{5}$^th of the previous height. The ball does not bounce further if the previous height is less than 50 cm. What is the number of times the ball hits the ground before the ball stops bouncing ?

The correct answer is Option 5 (or 5 times).

Step-by-step Explanation:

1. Initial height:
The ball is initially dropped from a height of 1.5 meters.
Let us convert meters into centimeters for easier calculation:
Height before the 1st bounce, $H_0$ = 1.5 m = 150 cm.

2. Condition for stopping:
The ball stops bouncing if the height attained after a bounce is less than 50 cm. That is, if the previous height before a potential bounce is less than 50 cm, it will not bounce further.

3. Calculating height after each bounce:

1st Hit (Bounce 1):
Height attained after the 1st hit, $H_1=\frac{4}{5}\times 150=120$ cm.
Since 120 cm ≥ 50 cm, the ball hits the ground again.

2nd Hit (Bounce 2):
Height attained after the 2nd hit, $H_2=\frac{4}{5}\times 120=96$ cm.
Since 96 cm ≥ 50 cm, the ball hits the ground again.

3rd Hit (Bounce 3):
Height attained after the 3rd hit, $H_3=\frac{4}{5}\times 96=76.8$ cm.
Since 76.8 cm ≥ 50 cm, the ball hits the ground again.

4th Hit (Bounce 4):
Height attained after the 4th hit, $H_4=\frac{4}{5}\times 76.8=61.44$ cm.
Since 61.44 cm ≥ 50 cm, the ball hits the ground again.

5th Hit (Bounce 5):
Height attained after the 5th hit, $H_5=\frac{4}{5}\times 61.44=49.152$ cm.
Now, after hitting the ground for the 5th time, the height attained is 49.152 cm.

4. Conclusion:
Since 49.152 cm is less than 50 cm, the ball does not bounce further. Thus, the total number of times the ball hits the ground before it stops bouncing is 5.