Answer
The initial speed is ${v_0} \simeq 67.28$ m/s.
Work Step by Step
Let ${v_0}$ denote the initial speed of the projectile. So, we have the initial velocity:
${\bf{v}}\left( 0 \right) = \left( {{v_0}\cos 60^\circ ,{v_0}\sin 60^\circ } \right) = \left( {\frac{1}{2}{v_0},\frac{1}{2}\sqrt 3 {v_0}} \right)$
Let the initial position be ${\bf{r}}\left( 0 \right) = \left( {0,0} \right)$.
The only force that acts on the projectile is the gravitational force. So, the acceleration due to gravity is ${\bf{a}}\left( t \right) = - 9.8{\bf{j}}$ $m/{s^2}$.
Step 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 ${\bf{v}}\left( 0 \right) = \left( {\frac{1}{2}{v_0},\frac{1}{2}\sqrt 3 {v_0}} \right)$ gives
$\left( {\frac{1}{2}{v_0},\frac{1}{2}\sqrt 3 {v_0}} \right) = \left( {0,0} \right) + {{\bf{c}}_0}$
${{\bf{c}}_0} = \left( {\frac{1}{2}{v_0},\frac{1}{2}\sqrt 3 {v_0}} \right)$
Thus,
${\bf{v}}\left( t \right) = \left( {0, - 9.8t} \right) + \left( {\frac{1}{2}{v_0},\frac{1}{2}\sqrt 3 {v_0}} \right)$
${\bf{v}}\left( t \right) = \left( {\frac{1}{2}{v_0},\frac{1}{2}\sqrt 3 {v_0} - 9.8t} \right)$
Step 2. Find the position vector
We have
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {\frac{1}{2}{v_0},\frac{1}{2}\sqrt 3 {v_0} - 9.8t} \right){\rm{d}}t$
${\bf{r}}\left( t \right) = \left( {\frac{1}{2}{v_0}t,\frac{1}{2}\sqrt 3 {v_0}t - 4.9{t^2}} \right) + {{\bf{c}}_1}$
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,
${\bf{r}}\left( t \right) = \left( {\frac{1}{2}{v_0}t,\frac{1}{2}\sqrt 3 {v_0}t - 4.9{t^2}} \right)$
Step 3. Solve for ${t_0}$
From previous result, we obtain the position vector:
${\bf{r}}\left( t \right) = \left( {\frac{1}{2}{v_0}t,\frac{1}{2}\sqrt 3 {v_0}t - 4.9{t^2}} \right)$
The projectile lands $400$ m away, so
$\frac{1}{2}{v_0}t = 400$ ${\ \ }$ and ${\ \ }$ $\frac{1}{2}\sqrt 3 {v_0}t - 4.9{t^2} = 0$
The first equation gives $t = \frac{{800}}{{{v_0}}}$. Substituting it in the second equation gives
$\frac{1}{2}\sqrt 3 {v_0}\left( {\frac{{800}}{{{v_0}}}} \right) - 4.9{\left( {\frac{{800}}{{{v_0}}}} \right)^2} = 0$
$400\sqrt 3 = \frac{{4.9 \times {{800}^2}}}{{{v_0}^2}}$
${v_0}^2 = \frac{{4.9 \times {{800}^2}}}{{400\sqrt 3 }} = 4526.43$
So, the initial speed is ${v_0} \simeq 67.28$ m/s.
