Chapter 3.6, Problem 56PS

Solution

To calculate: The height of the rider at the top of the swing if the rider’s maximum speed was 15 meters per second. Rider’s maximum speed v (in meters per second) occurs at the bottom of each swing and can be approximated by $v=\sqrt{2gh}$ where h is the height (in meters) at the top of each swing and g is the acceleration due to gravity $g=9.8\text{m}/{\sec }^2$ .

The height of the rider at the top of the swing is $11.48\text{}\text{m}$ .

Given information:

The equation is $v=\sqrt{2gh}$ representing the maximum speed of the rider at the bottom of each swing.

The given maximum speed is 15 meters per second and the acceleration due to gravity is $g=9.8\text{m}/{\sec }^2$ .

Formula used:

Raise each side of the equation to the same power.

Calculation:

Write the original equation:

  $v=\sqrt{2gh}$

Substitute the value of v as 15 and g as $9.8\text{m}/{\sec }^2$ into the equation,

  $15\frac{m}{\sec }=\sqrt{2\cdot 9.8\frac{m}{{\sec }^2}\cdot h}$

Square both sides of the equation,

  $\begin{array}{c}{\left(15\frac{m}{\sec }\right)}^2={\left(\sqrt{19.6\frac{m}{{\sec }^2}\cdot h}\right)}^2\\ 225\frac{m^2}{{\sec }^2}=19.6\frac{m}{{\sec }^2}\cdot h\end{array}$

Divide each side of the equation by $19.6\frac{m}{{\sec }^2}$ ,

  $\begin{array}{c}\frac{225\frac{m^2}{{\sec }^2}}{19.6\frac{m}{{\sec }^2}}=\frac{19.6\frac{m}{{\sec }^2}\cdot h}{19.6\frac{m}{{\sec }^2}}\\ h\approx 11.48m\end{array}$

Therefore, the height at the top of swing is $11.48m$ .