two students on a balcony 18.8m above the street. one student throws a ball vertically DOWNWARD 13.1 m/s; at the same instant, the onther student throws a ball vertically UPWARD at the same speed.
I. what is the volcity of the first ball as it strikes the ground? assume it is in the posotive direction... what is the volcity of the second ball as it strikes the ground?
$$g=9.81 \frac{m}{s^2} \quad v_0 = 13.1\frac{m}{s} \quad h=18.8\;m$$
$$\small{
\text{Way vertically downward: } s_1 = \frac{g}{2}t^2 \quad \text{and way thrown vertically downward or upward: } {s_2=v_0t\quad
}$$
$$\small{
way \;first\; ball\; s_I:\boxed{s_I= s_1+s_2 = h} \quad way \;second\; ball s_{II}:\boxed{s_{II}= s_1-s_2 = h}
}$$
$$\begin{array}{rcl|rcl}
\small{ \text{time first ball:} \ t_1 } &&& \small{ \text{time second ball:} \ t_2 }\\
s_I= \frac{g}{2}t_1^2+v_0t_1 & = & h
& s_{II}=\frac{g}{2}t_2^2-v_0t_2 &=&h\\
\frac{g}{2}t_1^2+v_0t_1 -h& = & 0
& \frac{g}{2}t_2^2-v_0t_2 -h&=& 0 \\
t_1>0: \quad t_1 &=& \frac{-v_0+\sqrt{v_0^2+2gh} }{g}
&t_2>0: \quad t_2 &=& \frac{v_0+\sqrt{v_0^2+2gh} }{g} \\
\end{array}$$
$$\begin{array}{rcl|rcl}
\small{ \text{final speed first ball:} \ v_1 } &&& \small{ \text{final speed second ball:} \ v_2 }\\
v_1&=& gt_1+v_0
& v_2 &=& gt_2-v_0 \\
v_1 & = & -v_0 + \sqrt{v_0^2+2gh} + v_0
& v_2 &=& v_0 + \sqrt{v_0^2+2gh} - v_0 \\
\end{array}$$
$$\boxed{v_1=v_2=\sqrt{v_0^2+2gh} } = \sqrt{13.1^2+2*9.81*18.8} = 23.2479246386\ \frac{m}{s}$$
II. what is the difference in the time the b***s spend in the air?
$$\begin{array}{rcl}
\Delta t & = & t_2-t_1 \\
&=& \frac{(v_0 + \sqrt{v_0^2+2gh})}{g} - \frac{( -v_0 + \sqrt{v_0^2+2gh} ) } {g}\\
&=& \frac{2v_0}{g}
\end{array}$$
$$\boxed{\Delta t=\frac{2v_0}{g} } = \frac{2*13.1}{9.81} = 2.67074413863\ s$$
III. how far apart are the b***s 0.964s after they are thrown?
$$\begin{array}{rcl}
\Delta s & = & s_I-s_{II} \\
&=& (s_1 + s_2 ) - ( s_1 - s_2 )\\
&=&2s_2 \\
&=&2v_0t
\end{array}$$
$$\boxed{
\Delta s=2v_0t } = 2*13.1*0.964
= 25.2568\ m$$
.
