Question

A wrecking ball on a 15.4 m longcable is pulled back to an angle of33.5° and released. At what speedis it moving at the bottom of itsswing?(Unit = m/s)Enter

Answer 1

Given data

*The given length of the cable is L = 15.4 m

*The given angle is

`\theta=33.5^0`

The formula for the height is given as

`h=L(1-\cos \theta)`

Substitute the values in the above expression as

`\begin{gathered} h=15.4(1-\cos 33.5^0) \\ =2.57\text{ m} \end{gathered}`

The formula for the velocity of the ball at the bottom of its swing is given as

`v=\sqrt[]{2gh}`

`\begin{gathered} v=\sqrt[]{2*9.8*2.57} \\ =7.09\text{ m/s} \end{gathered}`