a boll drop from a height 40 m then it come to ground and again jump its half of back

Its a G.P ques......a/(1-r)

where a=40 and r=1/2. So 40/(1- 0.5) + 20/(1-0.5)= 80+40= 120
8 years agoHelpfull: Yes(9) No(1)
S1=40+40(1/2)*2+40(1/2)^2*2+........
S2=40+2{ 40(1/2) + 40(1/2)^2 + 40(1/2)^3 +......+infinite term.}
here a=40(1/2) and r=1/2
Sum of infinite GP series S2=a/(1-r) = 40*(1/2) /1-(1/2)=40.
S1=80
so 40+2*40=120
8 years agoHelpfull: Yes(2) No(0)
40+20+20=80
8 years agoHelpfull: Yes(1) No(5)
Gp series question.... when ball first rebounce then its distance 4o m and then in 2nd rebounce half of 40 and its jump upside first and then go belowside so its distance becomes two times of half of 40 that is 40(1/2)*2. and similarly in third time rebounce 40(1/2)^2*2 and so on....
it's G.P series becomes 40 + 40(1/2)*2 + 40(1/2)^2*2 + 40(1/2)^3*2 + ........
40 + 2 { 40(1/2) + 40(1/2)^2 + 40(1/2)^3 +......+infinite term.}
40 + 2 * 40 { 1/2 + 1/2^2 + 1/2^3+1/2^4......+infinite term.}
here a = 1/2 and r=1/2
=40 + 80 ( 1/2/1-1/2) ( sum of infinite terms of G.P series is a/1-r ) [ a=1/2 and 1-r = 1-1/2=1/2 ]
=40+80
ans =120
8 years agoHelpfull: Yes(1) No(0)
80m= 40+20+20
8 years agoHelpfull: Yes(0) No(1)
The distance covered for the 1st time the ball goes up n come down is 40+40=80
the second time it goes upto half of 40 i.e. 20, but total dis now covered is 20+2==40 (both going up n back down)
so the series is like 80,40,20,10.......n so on
so, on applying G.P, a/(1-r) where a=80 n r=1/2 ,
the ans is 160
8 years agoHelpfull: Yes(0) No(0)

a boll drop from a height 40 m then it come to ground and again jump its half of back height then till it come to rest how it cover the distance from start point to its rest time..

optns nt remember