question_answer

A ball falls from a height of 2m and undergoes repeated collisions with the ground. Each time it hits the ground it raises to half of the height from which it fell. What is the time interval between third and fourth hits with the ground? What is the total distance travelled till the moment it makes fourth hit with the ground? Neglect the time of contact with the ground.

Answer:

Total distance = \[2+1+1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=5\frac{1}{2}m=5.5m\] Time between the third and fourth hits is time for moving 0.25 m up and 0.25 m down \[t=2\sqrt{\frac{2h}{g}}=2\sqrt{\frac{2\times \frac{1}{4}}{10}}=\frac{1}{\sqrt{5}}s\]