The projectile is fired with speed [tex]v[/tex] at an angle [tex]\theta[/tex] relative to the horizontal, so that the horizontal components of the initial velocity vector are
[tex]v_x=v\cos\theta[/tex]
[tex]v_y=v\sin\theta[/tex]
Its position vector has components
[tex]x=v\cos\theta\,t[/tex]
[tex]y=v\sin\theta\,t-\dfrac g2t^2[/tex]
where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex]
Recall that
[tex]{v_f}^2-{v_i}^2=2a\Delta x[/tex]
where [tex]v_i[/tex] and [tex]v_f[/tex] are the projectile's initial and final velocities, [tex]a[/tex] is its acceleration, and [tex]\Delta x[/tex] is its displacement.
In the vertical direction, velocity is 0 at maximum height and [tex]a=-g[/tex], so that
[tex]-\left(v\sin\theta\right)^2=-2gy_{\rm max}\implies y_{\rm max}=\dfrac{v^2\sin^2\theta}{2g}[/tex]
The projectile hits the ground when [tex]y=0[/tex], which happens for
[tex]0=v\sin\theta\,t-\dfrac g2t^2\implies t=\dfrac{2v\sin\theta}g[/tex]
where we omit [tex]t=0[/tex] because we know the projectile starts on the ground. This means the maximum horizontal range is
[tex]x_{\rm max}=v\cos\theta\left(\dfrac{2v\sin\theta}g\right)=\dfrac{2v^2\sin\theta\cos\theta}g[/tex]
We're given that [tex]x_{\rm max}=7y_{\rm max}[/tex], so
[tex]\dfrac{2v^2\sin\theta\cos\theta}g=\dfrac{7v^2\sin^2\theta}{2g}[/tex]
Solve for [tex]\theta[/tex]:
[tex]4\sin\theta\cos\theta=7\sin^2\theta[/tex]
[tex]\sin\theta(7\sin\theta-4\cos\theta)=0[/tex]
[tex]\implies\sin\theta=0\text{ OR }7\sin\theta-4\cos\theta=0[/tex]
The first case suggests that [tex]\theta=0[/tex], but then both the maximum height and range would be 0, which would technically satisfy the given condition, but it's not an interesting solution.
In the second case, we get
[tex]7\sin\theta-4\cos\theta=0[/tex]
[tex]7\sin\theta=4\cos\theta[/tex]
[tex]\tan\theta=\dfrac47[/tex]
[tex]\theta=\arctan\left(\dfrac47\right)\approx29.7^\circ[/tex]
So the projectile was launched at an angle of about 29.7º.
