Asoccer player kicks the ball toward a goal that is 16.8 m in front of him. the ball leaves his foot at a speed of 16.0 m/s and an angle of 28.08 above the ground. find the speed of the ball when the goalie catches it in front of the net.

final speed V = 14.71m/s

Explanation:

To determine the final speed

Given;

Distance of travel d= 16.8 m

Initial speed u = 16.0m/s

Angle to the horizontal a = 28.08°

Vx= ucosa = horizontal component of speed

Vy = usina = vertical component of speed

Time of travel t = d/Vx = 16.8m/16cos28.08 = 1.19s

The final vertical and horizontal component of speed

Vx = 16cos28.08 = 14.12m/s

Vy = Uy - gt = 16sin28.08 - 9.8(1.9) = -4.13m/s

Final speed

V = √(Vx^2 + Vy^2)

V = √( 14.12^2 + (-4.13)^2)

V = 14.71m/s

tbh i dont know ether

the components of a vector are the parts of the vectors that are to each other.

b perpendicular