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#
