Answer:
Exactly [tex]45^{\circ}[/tex] from the vertical ([tex]45^{\circ}[/tex] above the horizon,) assuming that air resistance on the projectile is negligible.
Explanation:
Let [tex]u_{x}[/tex] and [tex]u_{y}[/tex] denote the horizontal and vertical components of velocity of the projectile at launch, respectively. For the purpose of finding the kinetic energy of the projectile, the initial speed of the projectile would be:
[tex]\displaystyle u = \sqrt{{u_{x}}^{2} + {u_{y}}^{2}}[/tex].
Let [tex]m[/tex] denote the mass of this projectile. The initial kinetic energy of the projectile would be:
[tex]\displaystyle \frac{1}{2}\, m\, \left(\sqrt{{u_{x}}^{2} + {u_{y}}^{2}}\right)^{2} = \frac{1}{2}\, m\, \left({u_{x}}^{2} + {u_{y}}^{2}\right)[/tex].
Under the assumption that air resistance on the projectile is negligible, the horizontal component of velocity would stay the same ([tex]u_{x}[/tex]) during the entire flight.
However, in the vertical direction, the projectile would accelerate downward at [tex]a = (-g)[/tex] under the effect of gravitational attraction. Vertical component of velocity would change from upward to zero and then downward. When the projectile is at maximum height, the vertical component of the projectile would be exactly [tex]0[/tex]:
Hence, the kinetic energy of this projectile at maximum height would be:
[tex]\displaystyle \frac{1}{2}\, m\, \left(\sqrt{{u_{x}}^{2} + 0^{2}}\right)^{2} = \frac{1}{2}\, m\, {u_{x}}^{2}[/tex].
Given that the kinetic energy at maximum height is one-half the kinetic energy at launch:
[tex]\displaystyle \frac{1}{2}\, m\, {u_{x}}^{2} = \frac{1}{2}\, \left(\frac{1}{2}\, m\, \left({u_{x}}^{2} + {u_{y}}^{2}\right)\right)[/tex].
[tex]\displaystyle {u_{x}}^{2} = \frac{1}{2} \, \left({u_{x}}^{2} + {u_{y}}^{2}\right)[/tex].
[tex]\displaystyle {u_{x}}^{2} = {u_{y}}^{2}[/tex].
[tex]\displaystyle \frac{u_{y}}{u_{x}} = 1[/tex].
(Both vertical and horizontal component of initial velocity should be greater than zero.)
The angle of the initial velocity of this projectile can be found using the inverse trigonometric functions: