Explanation Given data: Initial velocity of skier is, v i = 12.0 m s . Angle between sloped section and the horizontal is, θ = 45 ∘ . Formula used: The formula for final displacement of the skier in x- direction is, x f = x i + ( v x ) i Δ t (1) x i is the initial displacement in x -direction, ( v x ) i is the x -component (horizontal) of initial speed of the skier, Δ t is the time duration of skier in the air. The formula for the final displacement of the skier in y - direction is, y f = y i + ( v y ) i Δ t − 1 2 g ( Δ t ) 2 (2) Here, y i is the initial displacement in y -direction, ( v y ) i is the y -component (vertical) of initial speed of the skier, g is the acceleration due to gravity, The expression for the vertical component of initial velocity ( y - direction ) is, ( v y ) i = v i sin θ (3) Here, v i is the initial velocity, θ is the angle of a downward sloped section with horizontal. The expression for the horizontal component of initial velocity ( x - direction ) is, ( v x ) i = v i cos θ (4) Explanation: The diagrammatic representation of given information is shown in figure 1, Substitute 45 ∘ for θ , and 12.0 m s for v i in equation (3) to find the vertical component of initial velocity, ( v y ) i = ( 12.0 m s ) sin 45 ∘ ( v y ) i = 8.485 m s Substitute 45 ∘ for θ , and 12.0 m s for v i in equation (4) to find the horizontal component of initial velocity, ( v x ) i = ( 12.0 m s ) cos ( 45 ∘ ) ( v x ) i = 8.485 m s From Figure 1, the initial displacement of skier in x – direction is, x i = 0 m . Therefore, substitute 0 m for x i , and 8.485 m s for ( v x ) i in equation (1) to find the expression for x f , x f = ( 0 m ) + ( 8

MODIFIED MASTERING COLLEGE PHYSICS 18WK.

4th Edition

ISBN: 9780136782216

Author: Knight

Publisher: PEARSON

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Chapter 3, Problem 60GP

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The time duration of skier in the air.

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Chapter 3, Problem 59GP

Chapter 3, Problem 61GP

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MODIFIED MASTERING COLLEGE PHYSICS 18WK.

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