Respuesta :
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
In that particular situation, you can prove it like this:
initial velocity is Vo
launch angle is α
initial vertical velocity is
Vv = Vo×sin(α)
horizontal velocity is
Vh = Vo×cos(α)
total time in the air is the the time it needs to fall back to a height of 0 m, so
d = v×t + a×t²/2
where
d = distance = 0 m
v = initial vertical velocity = Vv = Vo×sin(α)
t = time = ?
a = acceleration by gravity = g (= -9.8 m/s²)
so
0 = Vo×sin(α)×t + g×t²/2
0 = (Vo×sin(α) + g×t/2)×t
t = 0 (obviously, the projectile is at height 0 m at time = 0s)
or
Vo×sin(α) + g×t/2 = 0
t = -2×Vo×sin(α)/g
Now look at the horizontal range.
r = v × t
where
r = horizontal range = ?
v = horizontal velocity = Vh = Vo×cos(α)
t = time = -2×Vo×sin(α)/g
so
r = (Vo×cos(α)) × (-2×Vo×sin(α)/g)
r = -(Vo)²×sin(2α)/g
To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0.
dr/dα = d[-(Vo)²×sin(2α)/g] / dα
dr/dα = -(Vo)²/g × d[sin(2α)] / dα
dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα
dr/dα = -2 × (Vo)² × cos(2α) / g
Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when
cos(2α) = 0
2α = 90°
α = 45°
In that particular situation, you can prove it like this:
initial velocity is Vo
launch angle is α
initial vertical velocity is
Vv = Vo×sin(α)
horizontal velocity is
Vh = Vo×cos(α)
total time in the air is the the time it needs to fall back to a height of 0 m, so
d = v×t + a×t²/2
where
d = distance = 0 m
v = initial vertical velocity = Vv = Vo×sin(α)
t = time = ?
a = acceleration by gravity = g (= -9.8 m/s²)
so
0 = Vo×sin(α)×t + g×t²/2
0 = (Vo×sin(α) + g×t/2)×t
t = 0 (obviously, the projectile is at height 0 m at time = 0s)
or
Vo×sin(α) + g×t/2 = 0
t = -2×Vo×sin(α)/g
Now look at the horizontal range.
r = v × t
where
r = horizontal range = ?
v = horizontal velocity = Vh = Vo×cos(α)
t = time = -2×Vo×sin(α)/g
so
r = (Vo×cos(α)) × (-2×Vo×sin(α)/g)
r = -(Vo)²×sin(2α)/g
To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0.
dr/dα = d[-(Vo)²×sin(2α)/g] / dα
dr/dα = -(Vo)²/g × d[sin(2α)] / dα
dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα
dr/dα = -2 × (Vo)² × cos(2α) / g
Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when
cos(2α) = 0
2α = 90°
α = 45°
The maximum range of a projectile is at a launch angle of 45⁰.
Projectile motion
Projectile motion is the curved path of an object thrown into space under the influence of acceleration due to gravity.
The range is the horizontal displacement of a projectile. It is given by:
Range = (u²sin(2θ))/g
u is initial velocity, θ is angle, g is acceleration due to gravity.
Maximum range is at sin(2θ) = 1
2θ = sin⁻¹(1)
2θ = 90⁰
θ = 45⁰
The maximum range of a projectile is at a launch angle of 45⁰.
Find out more on Projectile motion at: https://brainly.com/question/24216590
