A football player kicks a football off a tee with a speed of 21 m/s at an angle of 51°. what is the horizontal component of the velocity?

The horizontal component of the velocity is approximately 13.22 m/s

Explanation:

The velocity of the kick which the football player gives the ball = 21 m/s

The angle of the path taken by the ball = 51°

The velocity can be represented vectorially, showing the vertical and horizontal components as follows;

v = 21 × cos(51°)· + 21 × sin(51°)·

Where we have;

The horizontal component of the velocity = 21 × cos(51°)

The vertical component of the velocity = 21 × sin(51°)

Therefore;

The horizontal component of the velocity = 21 × cos(51°) ≈ 13.22 m/s

The horizontal component of the velocity ≈ 13.22 m/s.

3.2x10^7