Question

3. A rescuer jumped from an airship in the ocean 1.20 x 102 m above the water's surface. Whatwas her kinetic energy at the moment she was 30.0 m from the water's surface? What was herspeed at that moment assuming her mass is 60.0 kg?

Answer 1

Given data,

The initial velocity of the body is zero.

The distance travelled by the rescuer upto the height of 30 m from the water surface is,

`\begin{gathered} S=102-30 \\ S=72\text{ m} \end{gathered}`

The final velocity of the rescuer at the height 30 m is,

`v^2-u^2=2gS`

where g is the acceleration due to gravity.

Substituting the known values,

`\begin{gathered} v^2=2*9.8*72 \\ v^2=1411.2 \\ v=37.6ms^2 \end{gathered}`

Thus, the kinetic energy of the rescuer is,

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

Substituting the known values,

`\begin{gathered} K=(1)/(2)*60*1411.2 \\ K=42336 \\ K=42.3\text{ KJ} \end{gathered}`

Thus, the kinetic energy of the rescuer is 42.3 KJ and speed of the rescuer is 37.6 meter per second square.