Chapter 5.4, Problem 14E

To determine

To calculate: the height at the top of the swing of the rider whose maximum speed is 15 metres per second.

Answer to Problem 14E

The height at the top of the swing of the rider whose maximum speed is 15 metres per second is 11.47 metres.

Explanation of Solution

Given information: in an amusement park ride, a rider suspended by cables, swings back and forth from a tower. The maximum speed v (in metre per second) of the rider can be approximated by $v=\sqrt{2gh}$ where h is the height (in metre) and g is gravity ( $g\approx 9.8m/{\sec }^2$ ).

Substitute speed v, 15 metres per second, and gravity g , in the provided equation,

  $\begin{array}{l}v=\sqrt{2gh}\\ 15=\sqrt{2\left(9.8\right)h}\\ \sqrt{19.6h}=15\end{array}$

Square both sides,

  $\begin{array}{l}{\left(\sqrt{19.6h}\right)}^2={\left(15\right)}^2\\ 19.6h=225\end{array}$

Divide by 19.6 on both sides,

  $\begin{array}{l}\frac{19.6h}{19.6}=\frac{225}{19.6}\\ h=11.47\end{array}$

Hence, the height at the top of the swing of the rider whose maximum speed is 15 metres per second is 11.47 metres