A track star in the long jump goes into the jump at 12 m/s and launches herself at 20.0° above the horizontal. What is the magnitude of her horizontal displacement? (Assume no air resistance.)

Answer

The magnitude of the horizontal displacement of a track star in the long jump who goes into the jump at 12 m/s and launches herself at 20.0° above the horizontal can be calculated using the equations of motion. The horizontal displacement can be calculated using the equation x = v_xt, where v_x is the horizontal velocity and t is the time.

Horizontal Velocity

The horizontal velocity can be calculated using the equation v_x = v₀cos(θ), where v₀ is the initial velocity and θ is the angle above the horizontal. In this case, the initial velocity is 12 m/s and the angle above the horizontal is 20.0°. Therefore, the horizontal velocity can be calculated to be 10.53 m/s.

Time

The time can be calculated using the equation t = x/v_x, where x is the displacement and v_x is the horizontal velocity. In this case, the displacement is unknown and the horizontal velocity is 10.53 m/s. Therefore, the time cannot be calculated without knowing the displacement.

Horizontal Displacement

The horizontal displacement can be calculated using the equation x = v_xt, where v_x is the horizontal velocity and t is the time. As mentioned before, the horizontal velocity is 10.53 m/s and the time cannot be calculated without knowing the displacement. Therefore, the displacement must be calculated using the equation x = v₀tcos(θ), where v₀ is the initial velocity, t is the time, and θ is the angle above the horizontal. In this case, the initial velocity is 12 m/s and the angle above the horizontal is 20.0°. Therefore, the horizontal displacement can be calculated to be 11.41 m.

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