Explanation Given data: The distance between still acrobat and jumped acrobat is 9 m. The speed of the acrobat leaving trampoline is 8 m s . Formula used: Consider the expression for final velocity. ( v y ) f 2 − ( v y ) i 2 = 2 a y Δ y (1) Here, ( v y ) i is the initial velocity, Δ t is the change in time, and a y is the acceleration. Consider the expression for distance. Δ y = ( v y ) i Δ t + 1 2 a y ( Δ t ) 2 (2) Here, ( v y ) i is the initial velocity, Δ t is the change in time, and a y is the acceleration. Calculation: Draw the diagram as shown in Figure 1. Modify equation (1) in terms of free fall acceleration. ( v y ) f 2 − ( v y ) i 2 = − 2 g Δ y (3) Here, g is the acceleration due to gravity which has a standard value of 9.8 m s 2 . Substitute 0 m s for ( v y ) f , 9.8 m s 2 for g and 8 m s for ( v y ) i ,in equation (3), ( 0 m s ) 2 − ( 8 m s ) 2 = − 2 × 9.8 m s 2 × y trampoline and acrobat ( 8 m s ) 2 = 2 × 9.8 m s 2 × y trampoline and acrobat y trampoline and acrobat = 64 m 2 s 2 2 × 9.8 m s 2 y trampoline and acrobat = 3.26 m So, the maximum height of the jumping acrobat from the trampoline is 3.26 m. So the total distance covered by the ball to reach the trampoline is, y total = y trampoline and acrobat + y distance between acrobat Substitute 3.26 m for y trampoline and acrobat and 9 m for y distance between acrobat , y total = 3.26 m + 9 m = 12 .26 m Consider y total as Δ y and the ball is starting from rest. Modify equation (2) in terms of acceleration due to gravity, Δ y = ( v y ) i Δ t + 1 2 g ( Δ t ) 2 (4) Substitute 12.26 m for Δ y , 9.8 m s 2 for g and 0 m s for ( v y ) i , 12

4th Edition

ISBN: 8220106755235

Author: Field

Publisher: PEARSON

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Question

Chapter 2, Problem 49P

To determine

The time period of the acrobat in air before he catches the ball.

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Chapter 2, Problem 48P

Chapter 2, Problem 50P

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