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 $t$.
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*}