In this problem, the mass of the mountain climber in Guided Example 6.11 was increased to $m = 91.0~mathrm{kg}$. We calculate what happens to the height $h$. We use $g = 9.81~mathrm{m/s^{2}}$.

From the example, we have

$$
begin{align*}
mgh &= PE_text{gravity} \
implies h &= left( frac{PE_text{gravity}}{g} right) frac{1}{m} tag{1} \
h &propto frac{1}{m}
end{align*}
$$

Since the product of $m$ and $h$ is constant, their relationship is inverse. If $m$ is increased, that means that $h$ would $textbf{decrease}$.

To calculate the altitude, we use equation (1) with $PE_text{gravity} = 8.87 times 10^{5}~mathrm{J}$ and $m = 91.0~mathrm{kg}$.

$$
begin{align*}
h &= left( frac{PE_text{gravity}}{g} right) frac{1}{m} \
&= left( frac{8.87 times 10^{5}~mathrm{J}}{9.81~mathrm{m/s^{2}}} right) frac{1}{91.0~mathrm{kg}} \
&= 993.60375~mathrm{m} \
h &= boxed{ 994~mathrm{m} }
end{align*}
$$

In this problem, we are given that a ball of mass $m = 0.25~mathrm{kg}$ is at an elevation $h = 1.3~mathrm{m}$ above the floor. We calculate its gravitational potential energy. We use $g = 9.81~mathrm{m/s^{2}}$.

From the definition, we have

$$
begin{align*}
PE_text{gravity} &= mgh \
&= left( 0.25~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right) left( 1.3~mathrm{m} right) \
&= 3.18825~mathrm{J} \
PE_text{gravity} &= boxed{3.2~mathrm{J}}
end{align*}
$$

$$
PE_text{gravity} = 3.2~mathrm{J}
$$

In this problem, the gravitation potential energy of a person on a diving board $h = 3.0~mathrm{m}$ high is $PE_text{gravity} = 1800~mathrm{J}$. We calculate the person’s mass. We use $g = 9.81~mathrm{m/s^{2}}$.

From the definition of gravitational potential energy, we have

$$
begin{align*}
PE_text{gravity} &= mgh \
implies m &= frac{PE_text{gravity}}{gh} \
&= frac{1800~mathrm{J}}{left( 1800~mathrm{J} right) left( 3.0~mathrm{m} right)} \
&= 61.16208~mathrm{kg} \
m &= boxed{ 61~mathrm{kg} }
end{align*}
$$

$$
m = 61~mathrm{kg}
$$