The equation for the conservation of mechanical energy is

$$
KE_text{i} + PE_text{i} = KE_text{f} + PE_text{f}
$$

All of the terms in the equation have a factor of $m$. Hence, the relationship of the speeds and height are independent of the mass. The speed is **the same** as the value calculated in the example.

In this problem, a player wins a point by hitting the ball of mass $m = 0.059~mathrm{kg}$ to the ground on the opponent’s side. The ball bounces with speed $v_text{i} = 16~mathrm{m/s}$ and a fan at an elevation $h$ catches it when it has a speed of $v_text{f} = 12~mathrm{m/s}$. We calculate how high ($h$) above the court is the fan. We use $g = 9.81~mathrm{m/s^{2}}$.

We use the equation for the conservation of mechanical energy. The potential on the ground is $0$, so we have

$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
frac{1}{2}mv_text{i}^{2} + 0 &= frac{1}{2}mv_text{f}^{2} + mgh \
implies h &= frac{v_text{i}^{2} – v_text{f}^{2}}{2g} \
&= frac{left( 16~mathrm{m/s} right)^{2} – left( 12~mathrm{m/s} right)^{2}}{2left( 9.81~mathrm{m/s^{2}} right)} \
&= 5.70846~mathrm{m} \
h &= boxed{5.7~mathrm{m}}
end{aligned}
$$

$$h = 5.7~mathrm{m}$$

In this problem, a crow drops a clam of mass $m = 0.11~mathrm{kg}$ onto a rocky beack from a height of $h_text{i} = 9.8~mathrm{m}$. We calculate the kinetic energy when the clam is at height $h_text{f} = 5.0~mathrm{m}$ and the speed of the clam at this point. We use $g = 9.81~mathrm{m/s^{2}}$.

The initial speed is $0$, so the kinetic energy is initially $0$. The equation of conservation of mechanical energy is

$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
0 + mgh_{1} &= KE_text{f} + mh_{2} \
implies KE_text{f} &= mg left( h_{1} – h_{2} right) \
&= left( 0.11~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right) left( 9.8~mathrm{m} – 5.0~mathrm{m} right) \
&= 5.17968~mathrm{J} \
KE_text{f} &= boxed{ 5.2~mathrm{J} }
end{aligned}
$$

To find the speed, we use the definition of the kinetic energy.

$$
begin{aligned}
KE_text{f} &= frac{1}{2}mv_text{f}^{2} \
v_text{f}^{2} &= frac{2KE_text{f}}{m} \
implies v_text{f} &= sqrt{frac{2KE_text{f}}{m}} \
&= sqrt{frac{2left( 5.17968~mathrm{J} right)}{0.11~mathrm{kg}}} \
&= 9.70443~mathrm{m/s} \
v_text{f} &= boxed{ 9.7~mathrm{m/s} }
end{aligned}
$$

$$
begin{aligned}
KE_text{f} &= 5.2~mathrm{J} \
v_text{f} &= 9.7~mathrm{m/s}
end{aligned}
$$