A soccer ball is kicked with an initial speed of 19.6 m/s in a direction 23.0° above the horizontal. Find the magnitude and direction of its velocity at the following times. (Take the +x axis to be parallel to the ground, in the direction that the ball is kicked.)

(a) 0.250 s after being kicked
m/s
° (counterclockwise from the +x axis)

(b) 0.500 s after being kicked
m/s
° (counterclockwise from the +x axis)

(c) Is the ball at its greatest height before or after 0.500 s?
---Select--- before after

To solve this problem, we can use the equations of motion for projectile motion. The key parameters we have are the initial speed (19.6 m/s) and the launch angle (23.0°).

(a) To find the magnitude and direction of the velocity at 0.250 s, we need to find the horizontal and vertical components of the velocity at that time.

The horizontal component of the velocity remains constant throughout the motion, so it will be the same as the initial horizontal velocity. It can be calculated using the equation:
Vx = V0 * cosθ

Substituting the given values:
Vx = 19.6 m/s * cos(23.0°)
Vx ≈ 17.7 m/s

The vertical component of the velocity changes due to the acceleration of gravity. We can calculate the vertical velocity at any time using the equation:
Vy = V0 * sinθ - gt

Substituting the given values:
Vy = 19.6 m/s * sin(23.0°) - 9.8 m/s^2 * 0.250 s
Vy ≈ 7.48 m/s

To find the magnitude of the velocity at 0.250 s, we use the Pythagorean theorem:
V = sqrt(Vx^2 + Vy^2)

Substituting the calculated values:
V = sqrt((17.7 m/s)^2 + (7.48 m/s)^2)
V ≈ 19.1 m/s

The direction of the velocity can be found using the inverse tangent function:
θ = atan(Vy / Vx)

Substituting the calculated values:
θ = atan(7.48 m/s / 17.7 m/s)
θ ≈ 23.9° (counterclockwise from the +x axis)

Therefore, the magnitude and direction of the velocity at 0.250 s are approximately 19.1 m/s and 23.9° counterclockwise from the +x axis, respectively.

(b) To find the magnitude and direction of the velocity at 0.500 s, follow the same steps as in part (a) but substitute the time t = 0.500 s.

Vx = 19.6 m/s * cos(23.0°)
Vx ≈ 17.7 m/s

Vy = 19.6 m/s * sin(23.0°) - 9.8 m/s^2 * 0.500 s
Vy ≈ 0 m/s (approximately zero since the ball reaches its peak and starts falling back down)

V = sqrt((17.7 m/s)^2 + (0 m/s)^2)
V ≈ 17.7 m/s

θ = atan(0 m/s / 17.7 m/s)
θ ≈ 0° (since Vx is positive)

Therefore, the magnitude and direction of the velocity at 0.500 s are approximately 17.7 m/s and 0° counterclockwise from the +x axis, respectively.

(c) To determine whether the ball is at its greatest height before or after 0.500 s, we can analyze the vertical motion.

At the highest point of the projectile, the vertical velocity is zero. From part (b), we found that at 0.500 s, the vertical velocity is approximately zero. This indicates that the ball reaches its maximum height before 0.500 s.

Therefore, the ball is at its greatest height before 0.500 s.