Chapter 11.2, Problem 19PP

To determine

The height that second diver has to leap to make a competitive splash.

Answer to Problem 19PP

The height that second diver has to leap to make a competitive splash is $1.1\text{ m}$ above the platform.

Explanation of Solution

Given:

The mass of the first diver is $m_1=136\text{ kg}$ .

The mass of the second diver is $m_2=100\text{ kg}$ .

The height of the platform is $h_1=3.0\text{ m}$ .

Formula used:

The expression for the gravitational potential energy is,

$PE=mgh$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $h$ is the distance from the reference level.

Calculation:

Consider the value of acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^2$ . If water is considered as a reference level, the potential energy will be equal to kinetic energy of the water just before hitting the water surface.

The potential energy of the first diver is,

$\begin{array}{l}PE=m_{1}g{h}_1\\ PE=\left(\text{136 kg}\right)\times \left(9.8\text{m}/{\text{s}}^2\right)\times \left(3.0\text{ m}\right)\\ PE=4\times {10}^3\text{ J}\end{array}$

The height of the platform that second diver have to jump is,

$\begin{array}{l}PE=m_{2}g{h}_2\\ 4\times {10}^3\text{ J}=\left(100.0\text{ kg}\right)\times \left(9.8\text{m}/{\text{s}}^2\right)\times h_2\\ h_2=4.10\text{ m}\end{array}$

The height would the second diver have to leap to make a competitive splash is,

$\begin{array}{l}h=h_2-h_1\\ h=4.1\text{ m}-3.0\text{ m}\\ h=1.1\text{ m}\end{array}$

Conclusion:

Thus, the height that second diver have to leap to make a competitive splash is $1.1\text{ m}$ above the platform.