Question

A soccer player kicks the ball toward a goal that is 29.0 m in front of him. The ball leaves his foot at a speed of 19.7 m/s and an angle of 30.6° above the ground. Find the speed of the ball when the goalie catches it in front of the net. (Note: The answer is not 19.7 m/s.)

________________ m/s

Answer 1

The horizontal component of the velocity is

v_x = (19.7m/s)cos(30.9^o) = 16.9 m/s

It remains constant at this value. The goal is 29 m from the point where the ball is kicked, so the time taken for the ball to reach the goal is

t = (29 m) / (16.9 m/s) = 1.72 s.

The vertical component of the velocity varies as

v_y = v_0y - gt

where the initial value is

v_oy = (19.7m/s)sin(30.6^o) = 10.02 m/s

After 1.72s,

v_y = 10.02 - (9.81)(1.72) = - 6.85 m/s

(The value is negative because the ball is falling as it approaches the goal.)

The speed is

v = sqrt{(16.9)² + (-6.85)²} = 18.24 m/s