Question

Still learning to dive, Simon steps off a 3.0 m high diving board with an initial velocity of zero. He falls straight down and hits the water feet first. Determine his velocity (m/s) at the moment of impact with the water. (neglect air resistance)

Answer 1

7.67 m/s

Explanation:

We can determine that the final velocity is given by the following equation:

`v^2=u^2+2\cdot a\cdot s`

In this case the acceleration is gravity, which would be gravity (9.8 m/s) and the distance (s), would be height.

We substitute and calculate the final speed:

`\begin{gathered} v^2=0^2+2\cdot9.8\cdot3 \\ v^2=0+58.8 \\ v=\sqrt[]{58.8} \\ v=7.67\text{ m/s} \end{gathered}`

The velocity at the moment of impact is 7.67 m/s