Question

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 1

We will have the following:

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

Now, we determine the velocity:

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

So, the velocity is 9.14 m/s.