Answer
The height $H$ at which the bullet hits the tower is $H = 355$ m.
Work Step by Step
We have the initial velocity:
${\bf{v}}\left( 0 \right) = \left( {120\cos \frac{\pi }{4},120\sin \frac{\pi }{4}} \right) = \left( {60\sqrt 2 ,60\sqrt 2 } \right)$
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 ${\bf{v}}\left( 0 \right) = \left( {60\sqrt 2 ,60\sqrt 2 } \right)$ gives
$\left( {60\sqrt 2 ,60\sqrt 2 } \right) = \left( {0,0} \right) + {{\bf{c}}_0}$
${{\bf{c}}_0} = \left( {60\sqrt 2 ,60\sqrt 2 } \right)$
Thus,
${\bf{v}}\left( t \right) = \left( {0, - 9.8t} \right) + \left( {60\sqrt 2 ,60\sqrt 2 } \right)$
${\bf{v}}\left( t \right) = \left( {60\sqrt 2 , - 9.8t + 60\sqrt 2 } \right)$
2. Find the position vector
We have
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {60\sqrt 2 , - 9.8t + 60\sqrt 2 } \right){\rm{d}}t$
${\bf{r}}\left( t \right) = \left( {60\sqrt 2 t, - 4.9{t^2} + 60\sqrt 2 t} \right) + {{\bf{c}}_1}$
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,
${\bf{r}}\left( t \right) = \left( {60\sqrt 2 t, - 4.9{t^2} + 60\sqrt 2 t} \right)$
3. Solve for $t$ when it hits the tower
From the previous result, we obtain the position vector:
${\bf{r}}\left( t \right) = \left( {60\sqrt 2 t, - 4.9{t^2} + 60\sqrt 2 t} \right)$
Since the tower is located $d=600$ m away, we solve for ${t_H}$ when it hits the tower:
$600 = 60\sqrt 2 {t_H}$
${t_H} = \frac{{10}}{{\sqrt 2 }}{\rm{s}}$
The height $H$ at which the bullet hits the tower is then
$H = - 4.9{t_H}^2 + 60\sqrt 2 {t_H}$
$H = - 4.9{\left( {\frac{{10}}{{\sqrt 2 }}} \right)^2} + 60\sqrt 2 \left( {\frac{{10}}{{\sqrt 2 }}} \right)$
$H = 355$ m
