A ball is thrown horizontally from the top of a 66.9 m building and lands 106.4 m from the base of the building. Ignore air resistance. (Assume the ball is thrown in the +x direction. Choose up to be the +y direction from the top of the building.)

a) How long is the ball in the air?

b) What must have been the initial horizontal component of the velocity? (Indicate the direction with the sign of your answer.)
Answer in m/s.

c) What must have been the vertical component of the velocity just before the ball hits the ground? (Indicate the direction with the sign of your answer.)
Answer in m/s.

d) What is the final velocity vector of the ball just before it hits the ground?
- Magnitude: Answer in m/s.
- Direction: Answer in degrees counter-clockwise from the +x direction.

d) The magnitude of the final velocity vector of the ball just before it hits the ground is approximately 46.0 m/s, and its direction is approximately 52.3° counter-clockwise from the +x direction.

To solve this problem, we can use the equations of motion for projectile motion. We'll consider the vertical and horizontal components of the ball's motion separately.

Height of the building (h) = 66.9 m

Horizontal distance traveled (d) = 106.4 m

Acceleration due to gravity (g) = 9.8 m/s²

a) To find the time the ball is in the air, we can use the equation for the vertical motion:

h = (1/2)gt²

Substituting the values:

66.9 m = (1/2)(9.8 m/s²)t²

Simplifying the equation:

t² = (66.9 m * 2) / 9.8 m/s²

t² = 2 * (66.9 m / 9.8 m/s²)

t² = 13.65 s²

Taking the square root of both sides:

t ≈ 3.7 s

Therefore, the ball is in the air for approximately 3.7 seconds.

b) The initial horizontal velocity component remains constant throughout the motion since there is no horizontal acceleration. Therefore, the initial horizontal component of the velocity is equal to the horizontal distance traveled divided by the time:

The horizontal component of velocity (Vx) = d / t

Substituting the values:

Vx = 106.4 m / 3.7 s

Vx ≈ 28.81 m/s

The initial horizontal component of the velocity is approximately 28.81 m/s in the positive x-direction.

c) To find the vertical component of the velocity just before the ball hits the ground, we can use the equation for the vertical motion:

Vfy = Viy + gt

Since the ball is thrown horizontally, the initial vertical component of velocity (Viy) is 0 m/s.

Vfy = 0 m/s + (9.8 m/s²)(t)

Substituting the time (t) calculated earlier:

Vfy = 0 m/s + (9.8 m/s²)(3.7 s)

Vfy ≈ 36.26 m/s

The vertical component of the velocity just before the ball hits the ground is approximately 36.26 m/s in the negative y-direction.

d) The final velocity vector of the ball just before it hits the ground can be calculated using the Pythagorean theorem:

Vf = sqrt((Vfx)² + (Vfy)²)

Since the ball was thrown horizontally, the final horizontal component of velocity (Vfx) is equal to the initial horizontal component of velocity (Vx).

Vf = sqrt((Vx)² + (Vfy)²)

Substituting the values:

Vf = sqrt((28.81 m/s)² + (36.26 m/s)²)

Vf ≈ 46.0 m/s

The magnitude of the final velocity vector of the ball just before it hits the ground is approximately 46.0 m/s.

To find the direction in degrees counter-clockwise from the +x direction, we can use the inverse tangent function:

θ = atan(Vfy / Vfx)

θ = atan(36.26 m/s / 28.81 m/s)

θ ≈ 52.3°

Therefore, the final velocity vector of the ball just before it hits the ground has a magnitude of approximately 46.0 m/s and a direction of approximately 52.3° counter-clockwise from the +x direction.

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