In this problem, we try to solve part (b) of Guided Example 3.12 using the equation $v_text{f}^{2} = v_text{i}^{2} + 2aDelta x$. It is given that $v_text{i} = 0$, $x_text{i} = 0$, $x = 3.00~mathrm{m}$, and $a = 9.81~mathrm{m/s^{2}}$. We confirm that $v_text{f} = 7.67~mathrm{m/s}$.

From the given, $Delta x = x – x_text{i} = 3.00~mathrm{m}$. Using the given equation, we have

$$
begin{align*}
v_text{f}^{2} &= v_text{i}^{2} + 2aDelta x \
v_text{f} &= sqrt{v_text{i}^{2} + 2aDelta x} \
&= sqrt{ 0 + 2 left( 9.81~mathrm{m/s^{2}} right) left( 3.00~mathrm{m} right) } \
&= 7.67203~mathrm{m/s} \
v_text{f} &= boxed{ 7.67~mathrm{m/s} }
end{align*}
$$

This implies that the same final speed can be calculated in different approaches.

In this problem, we may use the results from the previous problem. However, we are now given that $x = 10.0~mathrm{m}$. The other quantities are $v_text{i} = 0$, $x_text{i} = 0$, and $a = 9.81~mathrm{m/s^{2}}$. We find $v_text{f}$.

From the given, $Delta x = x – x_text{i} = 10.0~mathrm{m}$. Using the given equation, we have

$$
begin{align*}
v_text{f}^{2} &= v_text{i}^{2} + 2aDelta x \
v_text{f} &= sqrt{v_text{i}^{2} + 2aDelta x} \
&= sqrt{ 0 + 2 left( 9.81~mathrm{m/s^{2}} right) left( 10.0~mathrm{m} right) } \
&= 14.00714~mathrm{m/s} \
v_text{f} &= boxed{ 14.0~mathrm{m/s} }
end{align*}
$$

$$
v_text{f} = 14.0~mathrm{m/s}
$$

In this problem, a package dropped from height $H$ lands with speed $V$. We calculate what happens to the landing speed if the package is dropped from height $2H$.

We use the equation $v_text{f}^{2} = v_text{i}^{2} + 2aDelta x$ with $v_text{i} = 0$ and $Delta x = H$. We have

$$
begin{align*}
v_text{f}^{2} &= v_text{i}^{2} + 2aDelta x \
v_text{f} &= sqrt{v_text{i}^{2} + 2aDelta x} \
&= sqrt{0 + 2gH } \
&= sqrt{2g} sqrt{H} \
v_text{f} &propto sqrt{H}
end{align*}
$$

We see that the landing speed is proportional to the square root of the dropping height. Since the dropping height is doubled, the speed must be scaled by a factor of $sqrt{(2)}$. Hence, the landing speed of the package dropped from height $2H$ is
$$
boxed{ sqrt{2} V }
$$

$$
sqrt{2} V
$$

In this problem, a group of students stepped off a bridge and hit the riven after time $t = 1.5~mathrm{s}$. We find the height of the bridge. We use $g = 9.81~mathrm{m/s^{2}}$

Since the students stepped off, $v_text{i} = 0$. We have

$$
begin{align*}
h &= v_text{i}t + frac{1}{2}at^{2} \
&= 0 + frac{g}{2}t^{2} \
&= frac{9.81~mathrm{m/s^{2}}}{2} left( 1.5~mathrm{s} right)^{2} \
&= 11.03625~mathrm{m} \
h &= boxed{ 11~mathrm{m} }
end{align*}
$$

$$
h = 11~mathrm{m}
$$