Question

A platform diver with the mass of 51 kg, sieps off a divingboard that is elevated to a height of 10 meters above thewater. Assuming no air resistance, what will be her kineticenergy be right before she hits the water?

Answer 1

Given,

The mass of the diver, m=51 kg

The height, 10 m

Let us assume that the initial velocity of the diver is u=0 m/s

From the equation of motion,

`v^2=u^2+2gh`

Where v is the velocity of the diver just before she hits the water and g is the acceleration due to gravity.

On substituting the known values,

`\begin{gathered} v^2=0+2*9.8*10 \\ \Rightarrow v=\sqrt[]{196} \\ =14\text{ m/s} \end{gathered}`

The kinetic energy of the diver is given by,

`E=(1)/(2)mv^2`

On substituting the known values,

`\begin{gathered} E=(1)/(2)*51*14^2 \\ =4998\text{ J} \end{gathered}`

Thus the kinetic energy of the diver just before she hits the water is 4998 J