Respuesta :
Answer:
D = (- 5 i ^ + 15j ^) yard D = 15.81 yard , θ = -71.6º
Explanation:
This is an exercise of adding vectors, a diagram of the displacements is shown in the attached
d₁ = 10 j ^ y
d₂ = 10 i ^ y
The distance on the x-axis between receiver and quarterback is
Dₓ = d₂ - x₂
Dₓ = 10 - 15
Dₓ = - 5 i ^
The distance on the axis and between the receiver and the quarterback is
[tex]D_{y}[/tex] = d₁ -y₂
D_{y} = 10- (-5)
D_{y} = 15 j^
At this point is where the quarterback must throw the ball
D = (- 5 i ^ + 15j ^) yard
In the form of module and angle
D = √ (Dₓ² + D_{y}²)
D = √ (5² + 15²)
D = 15.81 yard
tan θ = D_{y} / Dₓ
θ = tan⁻¹ D_{y} / Dₓ
θ = tan⁻¹ (-15/5)
θ = -71.6º
