The projectile is launched with an angle of pi/3 which is equivalent to 60^\circ.
The horizontal distance the projectile will travel is obtained for y=0. Remember that the equation for the distance in the y component is: y = v_0sin(\theta)t-0.5g t^2.
So:
\qquad 0=v_0\sin(\theta)t*-0.5 g t^{2}
If we solve for t in the above equation:
\qquad v_0\sin(theta)t = 0.5 g t^2
\qquad v_0\sin(theta) = 0.5 g t
\qquad (v_0\sin(theta))/(0.5g)=t
\qquad t = (2v_0sin\theta)/(g)
And, the equation for the distance in the x component is: x=v_0\cos(theta)t.
Now, if we replace t in this equation, we will have:
\qquad x_{max} = v_0\cos(theta)t
\qquad x_{max} = v_0\cos theta*(2v_0sin\theta)/(g)
\qquad x_{max} =(2*v_0^2*cos theta*sin theta)/(g)
Remember that 2costhetasintheta=sin2theta.
\qquad x_{max} =(v_0^2*sin 2theta)/(g)
Finally, your maximum distance would be:
\qquad x_{max} =(v_0^2*sin 2theta)/(g)
\qquad x_{max} =(7^2*sin(2*60))/(9.81) = 4.3257 m approx 4.33m
