A soccer player kicks the ball toward a goal that is 30.0 m in front of him. The ball leaves his foot at a speed of 19.5 m/s and an angle?

Refer to the diagram below.

Take g = 9.81 m/s²
Take all upward quantities to be positive.

Consider the horizontal motion (uniform velocity = 30.0 m/s²) :
Time taken, t = 30.0 / (19.5 cos30.0°) s = 1.78 s

Consider the vertical motion (uniform acceleration):
v = vₒ + at
v = 19.5 sin30.0° + (-9.81) × (1.78) m/s
Vertical final velocity, v = -7.71 m/s

Final speed = √[(19.5 cos30.0°)² + 7.71²] m/s = 18.6 m/s

Note: The ball is caught before returning to the ground.

Initial vertical velocity = 19.5 * sin 30 = 9.75 m/s
Initial horizontal velocity = 19.5 * cos 30 = 9.75 * √3 ≈ 16.9 m/s
Let’s use the following equation to determine the total time the ball is in the air.

d = v * cos θ * t
30 = 9.75 * √3 * t
t = 30 ÷ (9.75 * √3)
The total time is approximately 1.78 seconds. Let’s use the following equation to determine the vertical velocity at this time.

vf = vi – g * t
vf = 9.75 – 9.8 * (30 ÷ (9.75 * √3)

This is approximately -7.7 m/s. Let’s use the following equation to determine the final speed.

Speed = √(Vertical^2 + Horizontal^2)
Speed = √([9.75 – 9.8 * (30 ÷ (9.75 * √3)]^2 + [(9.75 * √3)^2])

This is approximately 18.5 m/s.

0. The goalie catches the ball so the velocity has to be 0. If it’s not zero then it would be a goal.