Answer
See explanations.
Work Step by Step
Step 1. Compare Equation (2) in the Exercise with Equation (1) and the initial conditions in Exercise 103. We can set up the following differential equation $\frac{d^2s}{dt^2}=-g$, with initial conditions: $\frac{ds}{dt}=v_0$ and $s=s_0$ at $t=0$
Step 2. Work with the above conditions: given $\frac{d^2s}{dt^2}=-g$, we can get $\frac{ds}{dt}=\int(-g)dt=-gt+C$ and $s=\int(-gt+C)dt=-\frac{1}{2}gt^2+Ct+D$, where $C, D$ are constants.
Step 3. At $t=0, \frac{ds}{dt}=v_0$, we have $-g(0)+C=v_0$ and $C=v_0$
Step 4. At $t=0, s=s_0$, we have $-\frac{1}{2}g(0)^2+v_0(0)+D=s_0$ and $D=s_0$
Step 5. Thus $s=-\frac{1}{2}gt^2+v_0t+s_0$ which is the same as equation (2).
