Question # A ball is dropped from a height of ( 20 mathrm{m} ) and rebounds with a velocity which is ( 3 / 4^{text {th }} ) of the velocity with which it hits the ground. What is the time interval between the first and second bounces

# A ball is dropped from a height of ( 20 mathrm{m} ) and rebounds with a velocity which is ( 3 / 4^{text {th }} ) of the velocity with which it hits the ground. What is the time interval between the first and second bounces

( left(g=10 m / s^{2}right) )

(1) 3 sec

(2) 4 sec

(3) 5 sec

(4) 6 sec

Solution

If the boll is doopped from height h above the goound then it reaches the ground with speed.

[

V=sqrt{2 times 9 times h}

]

If it rebound with ( (3 / 9)^{text {th }} ) of the velocity then the taken t to reach maximum height after first bounce is given by

[

t=3 / 4 frac{sqrt{29} h}{9}=frac{3}{4} sqrt{frac{2 times h}{g}}=frac{3}{4} sqrt{frac{2 times 20}{10}}=frac{3}{2} mathrm{Se}

]

the ball will take sqmetime to reach the ground Just before 'second rebounce Hence the time interval between first and second rebounce is 3 seconds.