Question

An American football receiver starts at the 20 yard-line, 15 yards to the right of the quarterback. He runs 10 yards forward and then turns to the right and runs 10 yards before catching the ball thrown by the quarterback. If the quarterback is standing on the 15yard-line, how far must he throw the ball?

Answer 1

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

          $D_{y}$ = 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