Infinite series in context: bouncing balls and nested squares Problem set B One question at a time

A super-ball is dropped from the top of a climbing wall 9 m high. After each bounce it rises to 70% of the height of the bounce before. Find the height it reaches after the 7th bounce.

• \(u_n=9(0.7)^n\), with \(r=0.7\)
• \(u_7=9(0.7)^7\)
• \(u_7=0.741\) m

A steel bead is dropped onto a metal plate and takes 3 s to fall. Each time it bounces it takes 60% of the time of the previous fall to rise, and the same time to fall back.

1. (a) Show that the total time of the motion is given by \(3+6(0.6)+6(0.6)^2+\dots\) seconds.

   • Fall: 3 s
   • First bounce: rise \(0.6\times3=1.8\), fall 1.8, total \(6(0.6)\)
   • Second bounce: \(6(0.6)^2\). The pattern gives \(3+6(0.6)+6(0.6)^2+\dots\)

2. (b) Find a formula for \(S_n\), the total time of the first n terms.

   • After the first term: \(u_1=3.6,\ r=0.6\), with \(n-1\) terms
   • \(S_n=3+\frac{3.6(1-0.6^{n-1})}{1-0.6}\)
   • \(S_n=12-9(0.6)^{n-1}\)

A steel bead is dropped onto a metal plate and takes 3 s to fall. Each time it bounces it takes 60% of the time of the previous fall to rise, and the same time to fall back. Find the time from the drop until the bead comes to rest.

• Geometric part: \(u_1=3.6,\ r=0.6\)
• \(S_\infty=\frac{3.6}{1-0.6}=9\)
• Add the first fall: \(3+9=12\) s

Given that \(x+\frac{x^2}{4}+\frac{x^3}{16}+\dots=2\), find the value of x.

• \(u_1=x,\ r=\frac{x}{4}\)
• \(\frac{x}{1-\frac{x}{4}}=2\)
• From the calculator, \(x=\frac{4}{3}=1.33\)
• \(r=0.333\), so \(|r|<1\)

A basketball is dropped from a height of 8 m. After each bounce it rises to 85% of the height of the bounce before. Find the total distance it travels before it comes to rest.

• First bounce height \(u_1=8\times0.85=6.8\), \(r=0.85\)
• Up and down: \(\frac{2\times6.8}{1-0.85}=90.7\)
• Add the drop: \(8+90.7=98.7\) m

A square tile has sides of 14 cm. The midpoints of its sides are joined to make a new square, and the two triangles at the bottom-left and top-right corners are shaded. The same is done in the new square, and again, as shown. Find the total shaded area after 6 stages.

• Square \(14^2=196\)
• Two corner triangles are \(\frac28\) of the square: \(u_1=49\)
• Each new square has half the area, so \(r=\frac12\)
• \(S_6=\frac{49\left(1-\left(\frac12\right)^6\right)}{1-\frac12}=96.5\)

A square tile has sides of 14 cm. The midpoints of its sides are joined to make a new square, and the two triangles at the bottom-left and top-right corners are shaded. The same is done for ever. Find the total shaded area.

• \(u_1=49,\ r=\frac12\)
• \(S_\infty=\frac{49}{1-\frac12}=98\)
• Check: half the square is 98 cm²

A ball is dropped and bounces. The total time of the first n terms of its motion is \(S_n=12-9(0.6)^{n-1}\) seconds. Find the least value of n for which \(S_n>11.5\).

• \(12-9(0.6)^{n-1}>11.5\)
• From the calculator (table or solver): the first whole n that works
• \(S_{6}=11.3\), which is not enough. \(S_{7}=11.6\), which is.
• n = 7