(a)

#### To determine

**To find:** The total traveled distance of the ball.

#### Answer

The total traveled distance by the ball is H(1+r)1−r.

#### Explanation

**Given:**

The ball falls from a height h and it rebounds to a height rh where 0<r<1.

**Calculation:**

Initially, the ball falls a distance of H.

After the ball rebounds a distance rH, then the falls is rH.

The ball rebounds a distance of r2H, then the falls is r2H.

Proceed in the similar way, the total traveled distance of the ball is shown below,

D=H+rH+rH+r2H+r2H+r3H+r3H+⋯=H+2rH+2r2H+2r3H+⋯=(∑n=0∞2H⋅rn)−H=(2H∑n=0∞rn)−H

Since the sum of the geometric series is ∑n=0∞rn=11−r.

D=(2H⋅11−r)−H=2H1−r−H=2H−H(1−r)1−r=2H−(H−rH)1−r

Simply further,

D=H+Hr1−r=H(1+r)1−r

Therefore, the total traveled distance of the ball is, H(1+r)1−r meters.

(b)

#### To determine

**To calculate:** The total time that the ball travels.

#### Answer

The total time that the ball travels is 2Hg(1+r1−r) seconds.

#### Explanation

**Given:**

The ball falls 12gt2 meters in t seconds.

**Calculation:**

The total time tn it takes to the fall is Hrn meters for all n.

The value of tn calculated as shown below,

12gtn2=Hrntn2=2Hgrntn=2Hg(r)n

The total time is calculated as shown below,

T=t0+2t1+t22+⋯=∑n=0∞22Hg(r)n−t0=2Hg∑n=0∞2(r)n−2Hg=2Hg(21−r)−2Hg

Simplify further and obtain the time.

T=2Hg(21−r−1)=2Hg(2−1+r1−r)=2Hg(1+r1−r)

Therefore, the total time that the ball travels 2Hg(1+r1−r) seconds.

(c)

#### To determine

**To find:** The time taken for the ball to come to rest.

#### Answer

The time vg(1+k1−k) is taken for the ball to come to rest.

#### Explanation

**Given:**

The ball strikes the surface with the velocity v it rebound with velocity −kv.

**Calculation:**

From the part (b) the ball decelerate from knv meters per second to 0 or accelerate from 0 to knv.

The time segment tn computed as follows,

gtn=knvtn=knvg

The total time calculated as shown below.

T=t0+2t1+2t2+⋯=∑n=0∞2vgkn−vg=vg(∑n=0∞2kn−1)

Since the sum of the geometric series is ∑n=0∞2kn=21−k.

T=vg(21−k−1)=vg(1+k1−k)

Therefore, the time vg(1+k1−k) is taken for the ball to come to rest.