What is the speed of a Olympic diver just before they hit the water if they jump from a 9.14 m (30)foot platform ? assume they dive straight down.A) 24.26 m/s B) 57.25 m/sC) 179.33 m/s D) 13.39 m/s

Answer :

We will have the following:

[tex]\begin{gathered} d=vt+\frac{1}{2}at^2 \\ \\ \Rightarrow9.14=(0)t+\frac{1}{2}(9.8m/s^2)t^2\Rightarrow t^2=\frac{9.14}{4.9} \\ \\ \Rightarrow t=\sqrt{\frac{457}{245}}\Rightarrow t\approx1.365762103...s \end{gathered}[/tex]

Now, we determine the velocity:

[tex]\begin{gathered} v_f=v_it+at \\ \\ \Rightarrow v_f=(0m/s)(\sqrt{\frac{457}{245}}s)+\frac{1}{2}(9.8m/s^2)(\sqrt{\frac{457}{245}}s)^2\Rightarrow v_f=\frac{457}{50}m/s \\ \\ \Rightarrow v_f=9.14m/s \end{gathered}[/tex]

So, the velocity is 9.14 m/s.

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