Question

a ball is dropped from a height h above the ground it loses 50% of its velocity on every impact with the ground before bouncing up what is the total distance travelled by the ball when it finally comes on the ground?

Answer 1

the answer is 1.67h

Answer 2

well this type of questions usually goes in exponential forms a kind of exponential decay because imagine a ball loosing its velocity half of it in every impact..so it will be bouncing till infinite time.. if u say..
so our answer will be in approximation..

Answer 3

well start it from here
ball is dropped from height h hence the velocity it will attain before hitting the ground is V & the relation is
V^2-0^2=2*10*h
hence h=V^2/20

this is the prime relation we will be using in entire solution..

think for the interval 2 when ball goes up after hitting the ground..
our initial velocity is now v/2..
now apply this equation.. to find the distance travelled when it goes up..
0^2-(V/2)^2=2*10*S1
we have S1=(V^2/20)*1/4 no the total distance travelled when it goes up & down in total is 2S1
so distance travelled in this interval is
2S1=h/2

similarly when u solve for the 2nd interval when ball hits the ground & rebouncd with speed V/4
u will get the distance travelled as h/8.. for next interval..h/32 for next h/128

we have to sum
h+(h/2+h/8+h/32+h/128.............) this sum goes to infinity as i told earlier
the term in the bracket is a geometric progression having first term as h/2 & common ratio as 1/4

hence the total distance travelled is
h+(2h/3)=5h/3 approximately equals to 1.67h