Question

"We might think that a ball that is dropped from a height of 15 feet and rebounds to a height 7/8 of its previous height at each bounce keeps bouncing forever since it takes infinitely many bounces. This is not true! We examine this idea in this problem. A. Show that a ball dropped from a height h feet reaches the floor in 14h−−√ seconds. Then use this result to find the time, in seconds, the ball has been bouncing when it hits the floor for the first, second, third and fourth times:"B. How long, in seconds, has the ball been bouncing when it hits the floor for the nth time (find a closed form expression)? time at nth bounce = C. What is the total time that the ball bounces for? total time =

Answer 1

Answer:

Total Time = 4.51 s

Step-by-step explanation:

Solution:

- It firstly asks you to prove that that statement is true. To prove it, we will need a little bit of kinematics:

                             y = v_o*t + 0.5*a*t^2

Where,   v_o : Initial velocity = 0 ... dropped

              a: Acceleration due to gravity = 32 ft / s^2

              y = h ( Initial height )

                             h = 0 + 0.5*32*t^2

                             t^2 = 2*h / 32

                             t = 0.25*√h   ...... Proven

- We know that ball rebounds back to 7/8 of its previous height h. So we will calculate times for each bounce:

$1st : 0.25*\sqrt{15}\\\\2nd: 0.25*\sqrt{15} + 0.25*\sqrt{15*\frac{7}{8} } + 0.25*\sqrt{15*\frac{7}{8} } = 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} }\\\\3rd: 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 2*0.25*\sqrt{15*(\frac{7}{8} })^2\\\\= 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2\\\\4th: 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2 + 2*0.25*\sqrt{15*(\frac{7}{8} })^3$

$= 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2 + 0.5*\sqrt{15*(\frac{7}{8} })^3$

- How long it has been bouncing at nth bounce, we will look at the pattern between 1st, 2nd and 3rd and 4th bounce times calculated above. We see it follows a geometric series with formula:

  Total Time ( nth bounce ) = Sum to nth ( $\frac{1}{2}*\sqrt{15*(\frac{7}{8})^(^i^-^1^) } - \frac{1}{4}*\sqrt{15}$)

- The formula for sum to infinity for geometric progression is:

                                   S∞ = a / 1 - r

Where, a = 15 , r = ( 7 / 8 )

                                   S∞ = 15 / 1 - (7/8) = 15 / (1/8)

                                   S∞ = 120

- Then we have:

                                  Total Time = 0.5*√S∞ - 0.25*√15

                                  Total Time = 0.5*√120 - 0.25*√15

                                  Total Time = 4.51 s