I guess that this question is related to trajectories.
We have the basic equation s=ut-gt^2/2 where s is vertical distance, u initial vertical speed, g the acceleration of gravity and t is time. If an object has an initial speed magnitude of u and an initial projection angle of ø, the vertical component of the speed or velocity is usinø. So we have s=utsinø-gt^2/2. The horizontal distance travelled is unaffected by gravity, so is utcosø.
If a projectile is launched from a height h, then this has to to be taken into account by replacing s with s-h. The time for the projectile to fall to the ground can be found naturally under gravity is when utsinø-gt^2/2+h=0; this is a quadratic in t. This quadratic can be written gt^2-2utsinø-2h=0. Or t^2-2utsinø/g-2h/g=0. Completing the square we get t^2-2utsinø/g+u^2sin^2ø/g^2=2h/g+u^2sin^2ø/g^2; (t-usinø/g)^2=2h/g+u^2sin^2ø/g^2.
When t=usinø/g we have the vertex of the parabola. This is T[max]. The angle ø can be replaced by a so that we have T[max]=usina/g.
When t=usina/g, s=usina(usina/g)-(1/2)gu^2sin^2(a)/g^2=u^2sin^2(a)/2g=H[max], when h=0.
PROOF OF TRAJECTORY EQUATION USING CALCULUS
s"=-g is a simple expression that the derivative, or rate of change of vertical speed, of an object under gravity is the acceleration of gravity in a negative direction. Integrate this and we get the rate of change of vertical height (that is, the speed) is -gt+k where k is a constant. When t=0, speed is u, an initial speed. So v=u-gt or s'=u-gt. Finally we have a second integration: s=ut-gt^2/2+c where c is a constant. When t=0, c=h, the initial position of the projectile, so s=ut-gt^2/2+h. In this equation u is the vertical component of a velocity, hence usinø.