Answer:

We can use the equations of motion to solve this problem. Since the motion is only in the x-direction, we can focus on the x-components of the velocity and displacement.

Initial velocity in x-direction = 10 m/s (positive direction)

Time elapsed, t = 5 s

Acceleration in x-direction, a = 0 m/s^2 (assuming no air resistance)

Using the equation, x = ut + (1/2)at^2, where x is the displacement in x-direction, u is the initial velocity in x-direction, a is the acceleration in x-direction, and t is the time elapsed, we get:

x = (10 m/s)(5 s) + (1/2)(0 m/s^2)(5 s)^2 = 50 m

Therefore, Alice throws the ball to a horizontal distance of 50 m.

Next, we need to find the height the ball is thrown. Since there is no information given about the vertical motion, we assume that the ball follows a parabolic trajectory and experiences a constant downward acceleration of 9.81 m/s^2 due to gravity. Using the equation, y = ut + (1/2)at^2, where y is the displacement in y-direction, u is the initial velocity in y-direction (which is zero in this case), a is the acceleration in y-direction, and t is the time elapsed, we get:

y = (1/2)(-9.81 m/s^2)(5 s)^2 = -122.625 m

The negative sign indicates that the displacement is downward. Therefore, the height the ball is thrown is 122.625 m.

Finally, we can calculate the y-component of the velocity after it hits the ground using the equation, v = u + at, where v is the final velocity in y-direction after hitting the ground, u is the initial velocity in y-direction (which is zero in this case), a is the acceleration in y-direction, and t is the time elapsed after hitting the ground (which is also 5 s). We get:

v = 0 + (-9.81 m/s^2)(5 s) = -49.05 m/s

Therefore, the y-component of the velocity after it hits the ground is 49.05 m/s (downward direction).

Explanation:

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