Q A ball dropped from H height and moves 80 of height each time Total distance covered is Option 1

4. 9H
nitially distance covered is H.
then 80% of H=4H/5 in moving up of ball then it will go down=4H/5.
again 80% of 4H/5*2
this will make a g.p. of infinite terms.
hence h+2(4H/5+4/5*4H/5+....)
=h+8H=9H
11 years agoHelpfull: Yes(15) No(2)
Ans. 5H
First time distance is H
Second time 80H/100=4H/5
similarly third time 80%of 4H/5=H(4^2)/(5^2)
and so on...
this will lead to infinite terms of geometric progression
i.e
H+4H/5+16H/25............
sum = H/(1-4/5)=5H
11 years agoHelpfull: Yes(8) No(22)
distance = H + 2*0.8*H + 2*0.82*H + 2*0.83*H........

= 2*H*(1 + 0.8 + 0.82 + 0.83......) - H = 2*H/(1 - 0.8) - H = 2*H/0.2 - H = 10*H - H

= 9*H.
11 years agoHelpfull: Yes(6) No(5)
ans-5H
1st case 8H/10
2nd case (8/10)^2*H
3rd case (8/10)^3*H....so on and produce infinite gp
formula is (a/1-r)
8H/10(1+8/10+(8/10)^2.....)
8H/10+8H/10(so on....)
now total distance is H+4H=5H
11 years agoHelpfull: Yes(3) No(5)
it's a geometrical progression with an infinite number of terms, and it doesn't sum to infinity, even in theory. Only thing need to be careful of is, the first drop H only happens once, the subsequent heights happen twice.

So distance = H + 2*0.8*H + 2*0.82*H + 2*0.83*H........

= 2*H*(1 + 0.8 + 0.82 + 0.83......) - H = 2*H/(1 - 0.8) - H = 2*H/0.2 - H = 10*H - H

= 9*H.
7 years agoHelpfull: Yes(0) No(0)

Q. A ball dropped from H height and moves 80% of height each time. Total distance covered is

Option
(1) 4H
(2) 5H
(3) 7H
(4) 9H