A projectile is fired up from the surface of the earth with initial velocity $(u_0, v_0)$ A projectile is fired up from the surface of the earth
with initial velocity $(u_0, v_0)$. Under the influence of constant vertical acceleration −g the projectile
reaches height $h_{max}$ and then falls back to earth. Neglecting air resistance, show that the
fraction of time during its trajectory that the projectile spends above height $h_1$ is $|v_1|/v_0$, where
$(u_1, v_1)$ is the projectile’s velocity vector at height $h_1$. Assume that $0 ≤ h_1 ≤ h_{max}$
I am having trouble solving this because I can't figure out the time the projectile is in the air without a function being given. Or would I not even need that to determine the time it is above $h_1$? Thanks for the help!
 A: Solving
$$h_1=v_0t_1-1/2gt_1 \, \,\,$$
for $t_1$, where $t_1$ represents the two time intervals from the initial firing to the moments in which the projectile reaches the height $h_1$ in its ascending and descending phase, we get the two solutions
$$t_1=\frac{v_0\pm\sqrt{v_0^2-2gh_1}}{g}$$
So the time interval between these two moments is 
$$ \frac{2 \sqrt{v_0^2-2gh_1}}{g}$$
On the other hand, solving $v_0t_0-1/2gt_0=0 \,\,\,\,$ for $t_0$,  we get the two solutions
$$t_0=0$$
$$t_0=\frac{2v_0}{g}$$
whose difference $2v_0/g \,\,$ is the  total time from the initial firing to the moment in which the projectile comes back to the surface of the earth. So the searched proportion is
$$\frac{ 2 \sqrt{v_0^2-2gh_1}/g}{ 2v_0/g} \\   =\frac{\sqrt{v_0^2-2gh_1}}{v_0}$$
Now note that since $v_1=v_0 -gt_1 \,\,\,$, substituting $t_1$ with $t_1=\left(v_0\pm\sqrt{v_0^2-2gh_1} \right)/g \,\,\,$, 
we can write
$$v_1=\pm \sqrt{v_0^2-2gh_1}$$ 
where the $\pm$ indicates that, at height $h_1$, the velocity is positive in the ascending phase of the projectile and negative in the descending one.
From the last two equations, we directly get that the fraction of time during its trajectory that the projectile spends above height $h1$ is $|v_1|/v_0 \,\,$.
