A ball of mass m is droped from a height h , losing x% of its energy(that it had before the respective drop) on each fall.Calculate how many times the ball will fall until it stops and how long it will take the ball to do that.(?)
(the ball bounces up and down)
(the ball bounces up and down)
1 Answer
This is what I get
Explanation:
Initial potential energy of ball
Energy lost in the first bounce
Therefore, for the second drop height
To calculate time
h=ut+1/2g t^2h=ut+12gt2
Assuming that the ball is dropped with initial velocity
h=0xxt_1+1/2g t_1^2h=0×t1+12gt21
=>t_1=sqrt((2h)/g)⇒t1=√2hg .........(2)
From (1) we get time
t_2=sqrt((2h(1-x/100))/g)t2= ⎷2h(1−x100)g
=>t_2=t_1(1-x/100)^(1/2)⇒t2=t1(1−x100)12 ........(3)
For third drop as
=>t_3=t_1(1-x/100)^(2/2)⇒t3=t1(1−x100)22 .......(4)
Theoretically ball will bounce
t_1(1+(1-x/100)^(1/2)+(1-x/100)^(2/2)+(1-x/100)^(3/2).....)
Now sum of infinite GP is given by the expression as
S_(oo)=a_1/(1-r)
wherea_1 is the first term andr is common ratio and|r|<1
Hence we get total time taken by the ball to come to rest
T=t_1xx 1/(1-(1-x/100)^(1/2))
T=sqrt((2h)/g)xx1/(1-(1-x/100)^(1/2))