# A ball falls from a height of 1 m, hits the ground, and rebounds with half its velocity just before impact. After rising it falls and hits the ground and again rebounds with half its velocity just...

A ball falls from a height of 1 m, hits the ground, and rebounds with half its velocity just before impact. After rising it falls and hits the ground and again rebounds with half its velocity just before impact and so on. The total distance traveled by ball till it comes to rest is ______

A) 2m

B) 5/3m

C) 4/3m

D) 5/4m

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### 1 Answer

Hello!

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**.