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6. A professional football punter kicks a football with an initial velocity \( \mathbf{v} = (16.0 \, \text{m/s})\mathbf{i} + (23.0 \, \text{m/s})\mathbf{j} \). Determine the horizontal and maximum vertical displacements.

A. \( A_x = 37.6 \, \text{m}, A_y = 29.8 \, \text{m} \)

B. \( A_x = 37.6 \, \text{m}, A_y = 39.5 \, \text{m} \)

C. \( A_x = 75.0 \, \text{m}, A_y = 27.0 \, \text{m} \)

D. \( A_x = 75.0 \, \text{m}, A_y = 39.5 \, \text{m} \)

Answer :

The maximum vertical displacement, Ay, can be calculated using the equation for vertical displacement in projectile motion is 39.5 m, so correct option is B.

Displacement refers to the change in position of an object from its initial position to its final position. It is a vector quantity and is typically represented by the symbol "Δx" or "d."

Displacement includes both magnitude (the distance traveled) and direction (the change in position).

In the context of the given question, the horizontal displacement (Ar) represents the change in the x-coordinate or position in the horizontal direction, while the vertical displacement (Ay) represents the change in the y-coordinate or position in the vertical direction.

These values describe the magnitude of the changes in position in each respective direction.

The correct answer is:

b. Ar = 37.6 m.

Ay = 39.5 m

The horizontal displacement, Ar, can be calculated by multiplying the horizontal component of the initial velocity, which is 16.0 m/s, by the time of flight. The time of flight can be found by dividing the maximum vertical displacement by the vertical component of the initial velocity. Thus, the horizontal displacement is:

Ar = (16.0 m/s) * (39.5 m / 23.0 m/s)

= 37.6 m

The maximum vertical displacement, Ay, can be calculated using the equation for vertical displacement in projectile motion, considering the initial vertical velocity and the time of flight:

Ay = (23.0 m/s) * (39.5 m / 23.0 m/s) - (0.5 * 9.8 m/s² * (39.5 m / 23.0 m/s)²) = 39.5 m

Therefore, the correct answer is b. Ar = 37.6 m.

Ay = 39.5 m.

To know more about motion, visit:

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