Final answer:
The speed of the soccer ball at the top of its trajectory when launched at 12 m/s at an angle of 40 degrees is 9.192 m/s, which is the horizontal component of the initial velocity made invariant by the absence of a vertical component at that point.
Step-by-step explanation:
To find out how fast a soccer ball is going at the top of its trajectory when it is launched at 12 m/s at an angle of 40 degrees, we need to consider the components of the initial velocity. The vertical component of the velocity will be zero at the top of the trajectory because gravity will have decelerated it to a stop before it starts to fall back down. Therefore, we only need to consider the horizontal component of the velocity, as this does not change during the flight (assuming air resistance is negligible).
To calculate the horizontal velocity component (Vx), we use the initial velocity (V) and the cosine of the launch angle (θ):
Vx = V * cos(θ)
For the given values:
Vx = 12 m/s * cos(40°) = 12 m/s * 0.766 = 9.192 m/s
Therefore, the speed of the soccer ball at the top of its trajectory is 9.192 m/s and is only due to the horizontal component, as the vertical component is zero at that point.