Respuesta :
Answer:
v = (78.0 i ^ - 70.27 j ^) m/s, v = 105 m / s , θ = 318º
Explanation:
We have a projectile launch problem, let's start by calculating the time it takes to get through the canyon
y = [tex]v_{oy}[/tex] t - ½ gt2
As the motorcyclist comes out horizontally, the speed he has is the horizontal speed (vox) and the initial vertical speed is zero (I go = 0)
y = 0 - ½ g t2
t = √ 2y / g
t = √ (2 252 /9.8)
t = 7.17 s
Let's calculate the vertical speed for this time
[tex]v_{y}[/tex] = [tex]v_{oy}[/tex] - gt
[tex]v_{y}[/tex] = 0 - gt
[tex]v_{oy}[/tex] = - 9.8 7.17
[tex]v_{oy}[/tex] = - 70.27 m / s
We can give the result in two ways
First:
v = (78.0 i ^ - 70.27 j ^) m / s
Second:
using the Pythagorean theorem and trigonometry
v² = vₓ² + [tex]v_{oy}[/tex] ²
v = √ [(78.0)² + (-70.42)²] = √ (11042.98)
v = 105 m / s
tan θ₁ = [tex]v_{y}[/tex] y / vₓ
tan θ₁ = -70.42 / 78.0
θ₁ = 42º
If we measure this angle from the positive direction of the x-axis counterclockwise
θ = 360 - θ₁
θ = 360 - 42
θ = 318º
