Kline calculus intuitive chapter 3 problem 14 I am starting to get frustrated as I am not able to comprehend these questions because they simply do not make sense to me. 

A body is dropped. At the instant when it has fallen $100$ feet, a second body is to be projected downward with an initial velocity which will enable it to catch up with the first body in $10$ seconds after the second body is released. What is the requisite initial velocity?

Here is the problem and answer to it: 
https://scienceanswers.wordpress.com/2012/09/24/problem-3-2-14-page-48/.
First what I do not understand is why is $t_0$ for the second body $2.5$ seconds. Shouldn't it be $0$ seconds if it didn't travel any distance? Second if we integrate $a=32$ we get $v(t)= 32t+C$. Why can't I say that a time $2.5$ seconds which is supposedly $t_0$ for the second body $v(t)=0$ and just solve for $C$. I am never sure how to set the parameters such as $v$, $s$ and $t$. 
 A: 
A body is dropped. At the instant when it has fallen $100$ feet, a second body is to be projected downward with an initial velocity which will enable it to catch up with the first body in $10$ seconds after the second body is released. What is the requisite initial velocity?

Notice that the time the second body is released depends on when the first body has fallen $100~\text{ft}$.  Since the first body falls from rest, the distance it falls in $t$ seconds is 
$$d = \frac{1}{2} gt^2$$
Since the distance is given in feet, we take $g = 32~\text{ft}/\text{s}^2$.  Hence, we can determine when the first body has fallen $100~\text{ft}$ and when the second body is released by solving the equation
$$100~\text{ft} = \frac{1}{2}\left(32~\frac{\text{ft}}{\text{s}^2}\right)t^2$$
Solving for $t$ yields 
\begin{align*}
200~\text{ft} & = \left(32~\frac{\text{ft}}{\text{s}^2}\right)t^2\\
\frac{200~\text{ft}}{\left(32~\dfrac{\text{ft}}{\text{s}^2}\right)t^2} & = t^2\\
\frac{25}{4}~\text{s}^2 & = t^2\\
\frac{5}{2}~\text{s} & = t
\end{align*}
where we discard the negative solution for $t$ since the first body cannot fall before it is dropped.  Hence, the second body is released $2.5~\text{s}$ after the first body is dropped.  
To determine when the second body catches the first, we must set the distances they have fallen equal to each other.  We know that $t$ seconds after the first body has dropped, it has fallen
$$d_1(t) = \frac{1}{2}\left(32~\frac{\text{ft}}{\text{s}^2}\right)t^2$$
Since the second body does not begin to fall until $2.5~\text{s}$ after the first body is dropped, the distance the second body falls $t$ seconds after the first body is dropped is
$$d_2(t) = 
\begin{cases}
0, & t < \dfrac{5}{2}~\text{s}\\
v_0\left(t - \dfrac{5}{2}~\text{s}\right) + \dfrac{1}{2}\left(32~\dfrac{\text{ft}}{\text{s}^2}\right)\left(t - \dfrac{5}{2}~\text{s}\right)^2, & t \geq \dfrac{5}{2}~\text{s}
\end{cases}
$$
where we have used the formula $$d(t) = v_0t + \frac{1}{2}at^2$$ for the velocity of an object with initial velocity $v_0$ and acceleration $a$, with $a = 32~\text{ft}/\text{s}^2$.  
Since the second body catches up to the first body $10~\text{s}$ after the second body is released, which is $10~\text{s} + 2.5~\text{s} = 12.5~\text{s}$ after the first body is released, we obtain 
\begin{align*}
d_1(12.5~\text{s}) & = d_2(12.5~\text{s})\\
\frac{1}{2}\left(32~\frac{\text{ft}}{\text{s}^2}\right)(12.5~\text{s})^2 & = v_0\left(12.5~\text{s} - 2.5~\text{s}\right) + \frac{1}{2}\left(32~\dfrac{\text{ft}}{\text{s}^2}\right)\left(12.5 - 2.5~\text{s}\right)^2\\
\frac{1}{2}\left(32~\frac{\text{ft}}{\text{s}^2}\right)(12.5~\text{s})^2 & = v_0\left(10.0~\text{s}\right) + \frac{1}{2}\left(32~\dfrac{\text{ft}}{\text{s}^2}\right)\left(10.0~\text{s}\right)^2\\
2500~\text{ft} & = v_0(10~\text{s}) + 1600~\text{ft}\\
900~\text{ft} & = v_0(10~\text{s})\\
90~\dfrac{\text{ft}}{\text{s}} & = v_0
\end{align*}
