Let [tex]v_i[/tex] be the initial speed (m/s), so that [tex]v_{i,y}=v_i\sin60^\circ=\frac{\sqrt3}2v_i[/tex] (also m/s) is the vertical component of the initial velocity vector.
Recall that
[tex]{v_f}^2-{v_i}^2=2a\Delta y[/tex]
where [tex]v_i[/tex] and [tex]v_f[/tex] are the initial and final velocities, [tex]a[/tex] is the acceleration, and [tex]\Delta y[/tex] is the vertical displacement. At its maximum height (which the question seems to say occurs after 4 seconds), the velocity is 0, and throughout its motion the projectile is under the influence of gravity so that [tex]a=-g\frac{\rm m}{\mathrm s^2}[/tex] and g = 9.80.
[tex]-\left(\dfrac{\sqrt3}2v_i\right)^2=-2gy_{\rm max}\quad\quad(*)[/tex]
The projectile's height [tex]y[/tex] (m) at time [tex]t[/tex] (s) is
[tex]y=v_{i,y}t-\dfrac g2t^2[/tex]
so that when t = 4 s, its height is
[tex]y_{\rm max}=2\sqrt3v_i-8g\quad\quad(**)[/tex]
Solve [tex](*)[/tex] and [tex](**)[/tex] for [tex]v_i[/tex], then solve for [tex]y_{\rm max}[/tex]:
[tex](*)\implies v_i=\sqrt{\dfrac{8gy_{\rm max}}3}[/tex]
[tex](**)\implies v_i=\dfrac{y_{\rm max}+8g}{2\sqrt3}[/tex]
Then
[tex]\sqrt{\dfrac{8gy_{\rm max}}3}=\dfrac{y_{\rm max}+8g}{2\sqrt3}[/tex]
[tex]\implies\dfrac{{y_{\rm max}}^2-16gy_{\rm max}+64g^2}4=0[/tex]
[tex]\implies(y_{\rm max}-8g)^2=0[/tex]
[tex]\implies y_{\rm max}=8g\approx\boxed{78.4\,\mathrm m}[/tex]
