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

A ball is dropped from a window 15 m above the ground. After each bounce it rises to 60% of the height of the bounce before. Find the height it reaches after the 8th bounce.

• \(u_n=15(0.6)^n\), with \(r=0.6\)
• \(u_8=15(0.6)^8\)
• \(u_8=0.252\) m

A ping-pong ball is dropped onto a table and takes 1.5 s to fall. Each time it bounces it takes 80% 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 \(1.5+3(0.8)+3(0.8)^2+\dots\) seconds.

   • Fall: 1.5 s
   • First bounce: rise \(0.8\times1.5=1.2\), fall 1.2, total \(3(0.8)\)
   • Second bounce: \(3(0.8)^2\). The pattern gives \(1.5+3(0.8)+3(0.8)^2+\dots\)

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

   • After the first term: \(u_1=2.4,\ r=0.8\), with \(n-1\) terms
   • \(S_n=1.5+\frac{2.4(1-0.8^{n-1})}{1-0.8}\)
   • \(S_n=13.5-12(0.8)^{n-1}\)

A ping-pong ball is dropped onto a table and takes 1.5 s to fall. Each time it bounces it takes 80% of the time of the previous fall to rise, and the same time to fall back. Find the time from the drop until the ball comes to rest.

• Geometric part: \(u_1=2.4,\ r=0.8\)
• \(S_\infty=\frac{2.4}{1-0.8}=12\)
• Add the first fall: \(1.5+12=13.5\) s

Given that \(x+\frac{x^2}{5}+\frac{x^3}{25}+\dots=4\), find the value of x.

• \(u_1=x,\ r=\frac{x}{5}\)
• \(\frac{x}{1-\frac{x}{5}}=4\)
• From the calculator, \(x=\frac{20}{9}=2.22\)
• \(r=0.444\), so \(|r|<1\)

A tennis ball is dropped from a balcony 20 m high. After each bounce it rises to 70% of the height of the bounce before. Find the total distance it travels before it comes to rest.

• First bounce height \(u_1=20\times0.7=14\), \(r=0.7\)
• Up and down: \(\frac{2\times14}{1-0.7}=93.3\)
• Add the drop: \(20+93.3=113\) m

A square has sides of 10 cm. The midpoints of its sides are joined to make a new square, and the four corner triangles that are left over are shaded. The same is done in the new square, and again, as shown. Find the total shaded area after 6 stages.

• Square \(10^2=100\)
• Four corner triangles are \(\frac48\) of the square: \(u_1=50\)
• Each new square has half the area, so \(r=\frac12\)
• \(S_6=\frac{50\left(1-\left(\frac12\right)^6\right)}{1-\frac12}=98.4\)

A square has sides of 10 cm. The midpoints of its sides are joined to make a new square, and the four corner triangles that are left over are shaded. The same is done for ever. Find the total shaded area.

• \(u_1=50,\ r=\frac12\)
• \(S_\infty=\frac{50}{1-\frac12}=100\)
• Check: the whole square is 100 cm²

A ball is dropped and bounces. The total time of the first n terms of its motion is \(S_n=13.5-12(0.8)^{n-1}\) seconds. Find the least value of n for which \(S_n>12\).

• \(13.5-12(0.8)^{n-1}>12\)
• From the calculator (table or solver): the first whole n that works
• \(S_{10}=11.9\), which is not enough. \(S_{11}=12.2\), which is.
• n = 11