Answer
The initial speed of the bullet must be $72.75$ m/s.
Work Step by Step
Let ${v_0}$ denote the initial speed of the bullet. So, we have the initial velocity:
${\bf{v}}\left( 0 \right) = \left( {{v_0}\cos 45^\circ ,{v_0}\sin 45^\circ } \right) = {v_0}\left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)$
Let the initial position be ${\bf{r}}\left( 0 \right) = \left( {0,0} \right)$.
The only force that acts on the bullet is the gravitational force. So, the acceleration due to gravity is ${\bf{a}}\left( t \right) = - 9.8{\bf{j}}$ $m/{s^2}$.
1. Find the velocity vector
We have
${\bf{v}}\left( t \right) = \smallint {\bf{a}}\left( t \right){\rm{d}}t = \smallint \left( {0, - 9.8} \right){\rm{d}}t = \left( {0, - 9.8t} \right) + {{\bf{c}}_0}$
The initial condition ${\bf{v}}\left( 0 \right) = {v_0}\left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)$ gives
${v_0}\left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right) = \left( {0,0} \right) + {{\bf{c}}_0}$
${{\bf{c}}_0} = {v_0}\left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)$
Thus,
${\bf{v}}\left( t \right) = \left( {0, - 9.8t} \right) + {v_0}\left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)$
${\bf{v}}\left( t \right) = \left( {\frac{{{v_0}}}{2}\sqrt 2 , - 9.8t + \frac{{{v_0}}}{2}\sqrt 2 } \right)$
2. Find the position vector
We have
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {\frac{{{v_0}}}{2}\sqrt 2 , - 9.8t + \frac{{{v_0}}}{2}\sqrt 2 } \right){\rm{d}}t$
${\bf{r}}\left( t \right) = \left( {\frac{{{v_0}}}{2}\sqrt 2 t, - 4.9{t^2} + \frac{{{v_0}}}{2}\sqrt 2 t} \right) + {{\bf{c}}_1}$
The initial condition ${\bf{r}}\left( 0 \right) = \left( {0,0} \right)$ gives
$\left( {0,0} \right) = \left( {0,0} \right) + {{\bf{c}}_1}$
${{\bf{c}}_1} = \left( {0,0} \right)$
Thus,
${\bf{r}}\left( t \right) = \left( {\frac{{{v_0}}}{2}\sqrt 2 t, - 4.9{t^2} + \frac{{{v_0}}}{2}\sqrt 2 t} \right)$
3. Solve for ${v_0}$
The top of a $120$-m tower located $180$ m away can be represented by the point $\left( {180,120} \right)$. Thus, the bullet hits this point if there exists a time $t$ such that
${\bf{r}}\left( t \right) = \left( {\frac{{{v_0}}}{2}\sqrt 2 t, - 4.9{t^2} + \frac{{{v_0}}}{2}\sqrt 2 t} \right) = \left( {180,120} \right)$
In components we get
$\frac{{{v_0}}}{2}\sqrt 2 t = 180$, ${\ \ \ }$ $ - 4.9{t^2} + \frac{{{v_0}}}{2}\sqrt 2 t = 120$
The first equation yields $t = \frac{{360}}{{{v_0}\sqrt 2 }}$. Substituting it in the second equation gives
$ - 4.9{\left( {\frac{{360}}{{{v_0}\sqrt 2 }}} \right)^2} + \frac{{{v_0}}}{2}\sqrt 2 \left( {\frac{{360}}{{{v_0}\sqrt 2 }}} \right) = 120$
$ - 4.9{\left( {\frac{{360}}{{{v_0}\sqrt 2 }}} \right)^2} + 180 = 120$
$4.9{\left( {\frac{{360}}{{{v_0}\sqrt 2 }}} \right)^2} = 60$
Solving for ${v_0}$ we obtain ${v_0} = 72.75$. Thus, the initial speed of the bullet must be $72.75$ m/s.
