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
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Given data
*The given length of the cable is L = 15.4 m
*The given angle is
[tex]\theta=33.5^0[/tex]The formula for the height is given as
[tex]h=L(1-\cos \theta)[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} h=15.4(1-\cos 33.5^0) \\ =2.57\text{ m} \end{gathered}[/tex]The formula for the velocity of the ball at the bottom of its swing is given as
[tex]v=\sqrt[]{2gh}[/tex][tex]\begin{gathered} v=\sqrt[]{2\times9.8\times2.57} \\ =7.09\text{ m/s} \end{gathered}[/tex]