Question

Until his seventies, Henri LaMothe (Figure 9-48.) excited audiences by belly-flopping from a height of 12 minto 30 cmof water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass, find the magnitude of the impulse on him from the water.

Short answer

The magnitude of the impulse from the water, $\mathrm{J}=1.1\times {10}^3\mathrm{kg}.\mathrm{m}/\mathrm{s}$

Step-by-step solution

Understanding the given information

The estimated mass of Henri,$\mathrm{m}=70\mathrm{kg}$

The height above the water,$h=12\mathrm{m}$

The depth of water, $\mathrm{d}=30\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.30\mathrm{m}$

Concept and Formula used for the given question

We can use the equation of impulse relating to the change in momentum to calculate the impulse of water. The speed of Henri at the surface of the water can be calculated using the kinematic equation which is given as.

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ \mathrm{P}=\mathrm{m}\left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ {\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_{\mathrm{i}}^2+2\mathrm{ay} \end{gathered}$$

Calculation for the magnitude of the impulse on him from the water

The kinematic equation

$v_f^2=v_i^2+2ay$

Using this equation to calculate the velocity of Henri when he touches the surface of the water, we get

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}={\left(0\mathrm{m}/\mathrm{s}\right)}^2+\sqrt{2\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 12} \\ =15.34\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, using the impulse-momentum theorem, the impulse by the water is,

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ =\mathrm{m}\times \left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ =70\mathrm{kg}\times \left(0\mathrm{m}/\mathrm{s}-15.34\mathrm{m}/\mathrm{s}\right) \\ =-1.1\times {10}^3\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the magnitude of impulse is,

$\mathrm{J}=1.1\times {10}^3\mathrm{kg}.\mathrm{m}/\mathrm{s}$