Step-by-step explanation:
The angle that the tangent direction to the trajectory forms with the horizon is given by
{eq}\phi = \arctan \left( - \dfrac{v_y}{v_x} \right) \, , {/eq}
where
{eq}v_x = v \cos \theta \, , \\ v_y = v \sin \theta - g t {/eq}
are the components of the projectile velocity along the radial and the vertical direction, respectively, and
{eq}\theta \in \{ -\dfrac{\pi}{2}, \dfrac{\pi}{2}\} \, {/eq}
is the initial angle formed by the projectile with respect to the horizon.
We are interested in minimizing the angle {eq}\phi {/eq} at the time of impact {eq}t_0 \, . {/eq}
The latter can be obtained from the the formula for the vertical displacement, which relates it to the initial height h
{eq}y = h + v_y \ t - \dfrac{1}{2} \ g \ t^2 {/eq}
and noticing that at the time of impact y = 0 , from which follows
t
0
=
1
g
(
v
y
±
√
2
g
h
+
v
2
y
)
.
The sign choice in {eq}t_0 {/eq} corresponds to the choice of diametrically opposite initial velocities: as they describe the same trajectory, the result will be independent on such choice. In the following, we will chose the positive sign (corresponding to a projectile moving upwards after the explosion, which has a longer time of flight with respect to the one moving upwards).
t
0
corresponds to the choice of diametrically opposite initial velocities: as they describe the same trajectory, the result will be independent on such choice. In the following, we will chose the positive sign (corresponding to a projectile moving upwards after the explosion, which has a longer time of flight with respect to the one moving upwards).
The angle of impact reads therefore
ϕ
=
arctan
(
√
2
g
h
+
(
v
sin
θ
)
2
)
v
cos
θ
.
All the variables in the expression for {eq}\phi {/eq} are fixed except for the initial angle with respect to the horizon {eq}\theta {/eq}. Minimization of {eq}\phi {/eq} with respect to this angle gives
ϕ
are fixed except for the initial angle with respect to the horizon
θ
. Minimization of
ϕ
with respect to this angle gives
∂
ϕ
∂
θ
=
v
sin
(
θ
)
√
2
g
h
+
v
2
sin
2
(
θ
)
=
0
→
θ
=
0
to which corresponds the minimal impact angle
ϕ
=
tan
−
1
(
√
2
g
h
v
)
.