Respuesta :
Answer:
a) x = v₀ₓ t , y = [tex]v_{oy}[/tex] t - ½ g t²
Explanation:
This is a projectile launch problem, let's use Newton's second law on each axis
X axis
F = m a
Since there is no acceleration on the x axis, the force on this axis is zero
Y Axis
-W = m [tex]a_{y}[/tex]
-m g = m [tex]a_{y}[/tex]
[tex]a_{y}[/tex] = -g
In this axis the acceleration is the acceleration of gravity
Now we can use science to find the position of the body on each axis
X axis
x = v₀ₓ t + ½ a t²
As the acceleration on this axis is zero
x = v₀ₓ t
Y Axis
y = [tex]v_{oy}[/tex] t + ½ [tex]a_{y}[/tex] t²
The acceleration on this axis is –g
y = [tex]v_{oy}[/tex] t - ½ g t²
B) to find the maximum value of distance r
r =√ x² + y²
r = √( v₀ₓ² t² + ([tex]v_{oy}[/tex] t + ½ g t²)²
We can find the maximum value of r using time respect derivatives
dr / dt = 0
0 = ½ 1/√( v₀ₓ² t² + ([tex]v_{oy}[/tex] t + ½ g t²)² (v₀ₓ² 2t + 2 ([tex]v_{oy}[/tex] t - ½ g t²)([tex]v_{oy}[/tex] - ½ g2t)
We simplify this expression
0 = v₀ₓ² 2t + 2 ([tex]v_{oy}[/tex] t - ½ g t²) ([tex]v_{oy}[/tex] - ½ g2t)
-v₀ₓ² t =( [tex]v_{oy}[/tex] t - ½ g t²) ([tex]v_{oy}[/tex] - ½ g2t)
-v₀ₓ² t = [tex]v_{oy}[/tex]² t -3/2 gt² [tex]v_{oy}[/tex] + 1/2 g² t³
½ g² t²- 3/2 g [tex]v_{oy}[/tex] t = [tex]v_{oy}[/tex]² + v₀ₓ²
Let's use trigonometry to find go and vox
sin θ = [tex]v_{oy}[/tex] / v₀
cos θ = vox / v₀
[tex]v_{oy}[/tex] = v₀ sin θ
v₀ₓ = v₀ cos θ
We replace
½ g² t² -3/2 g [tex]v_{oy}[/tex] t = v₀ (sin² θ + cos² θ)
g t² - 3v₀ sin θ t = 2 v₀/g
The time is maximum for the angle is zero
