Denote the starting height as a.
At the stage 0 the ball falls from the height a and travels the distance a. Let's find its speed when it almost hits the ground:
`v_0(t) = -g*t,` `h_0(t) = a - (g*t^2)/2.`
Ball hits the ground when `h_0(t)=0,` i.e. `t_0=sqrt((2a)/g)` and `v_0 = v(t_0)=-g*t_0=-sqrt(2ag).`
At the stage 1 the ball rebounds and springs up from the initial height 0 with the initial speed `v_1 = |v_0/2| = sqrt((ag)/2).`
The height at this stage is `h_1(t) = v_1*t - (g*t^2)/2,` the velocity is `v_1(t) = v_1 - g*t.`
Ball hits the ground when `h_1(t_1)=0,` t>0, i.e. `v_1 = (g*t_1)/2,` `t_1=(2*v_1)/(g) = sqrt((2a)/g).` The ending speed will be the same as the initial (except the sign).
The maximum height will be
`h_1(t_1/2) = sqrt((ag)/2)*sqrt(a/(2g)) - (g/2)*(a/(2g)) = a/2 - a/4 = a/4.`
and the distance travelled, `d_1` , will be twice this, `a/2.`
Also this distance may be expressed as
`d_1 = 2*h_1(t_1/2) = 2*h_1(v_1/g) = 2*[((v_1^2)/g) - (g/2)*(v_1^2/g^2)] = v_1^2/g.`
The step 2 is similar to the step 1, with the initial speed of `v_2=v_1/2.` Therefore the distance will be `d_2 = (v_2^2/g) = (1/4)*(v_1^2/g) = (1/4)*d_1.`
Similarly `v_3 = v_2/2 = v_1/4, d_3 = (v_3^2/g) = (1/16)*d_1.`
`v^4 = v_1/8, d_4 = (1/64)*d_1.`
The complete distance will be `a + d_1*sum_(n=0)^(oo)(1/2^n) = a + (a/2)*2 = a*2.`
a=1m, so the answer is 2 (m), option A.