Sections:

Two Dimensional Motion and Vectors, Page 8

Solved Examples of Finding Vector Components

  1. A baseball is hit with a velocity of 14 m/s at angle of 30° with the horizontal direction. What are the horizontal and vertical vector components of the velocity vector? 

    Solution: 

    Given: v = 14 m/s, horizontal component vx = ? and the vertical component vy = ?

    Since the velocity vector makes angle θ with the horizontal then

V sub x = V cosine of theta

and 

v sub y = v sine of theta

Substituting the values of V and θ = 30° we get,

v sub x = 14 x cos 30 degrees; v sub x = 14 x 0.8660; v sub x = 12.1 m/s

and the vertical component is 

v sub y = v sine of theta

v sub y = 14 x sin 30 degrees; v sub = 14 x 0.5; v sub y = 7 m/s

  1. A bullet shot from the gun at an angle of 60°. Its horizontal component is 40 m/s. What is the actual velocity? 

    Solution: 

    Given: θ = 60°, vx = 40 m/s and v = ?

    The horizontal component of the velocity. 

    v sub x = v cos theta; 40 = v cos 60 degrees; v = 40 / cos 60 degrees; v = 40 / 0.5; v = 80 m/s

  2. The horizontal and vertical components of the velocity vector of the foot ball are 10√3m/s and 10 m/s respectively. What is the angle at which the ball is thrown? 

    Solution: 

    Given: v sub x = 10 square root of 3 m/s and v sub y = 10 m/s

    We know, V sub x = v cos of theta and v sub y = v sine of theta. Dividing these two equations we get

v sub x / v sub y = v sine of theta / v cosine of theta; v sub x / v sub y = v tangent of theta; 10 / 10 square root of 3 = tangent of theta; theta = tan ^-1 (1/square root of 3); theta = 30 degrees