Explanation Given information: The acceleration ( a E ) at a rate of 1.2 m/s 2 ( ↑ ) until it reaches a speed ( v E ) of 7.8 m/s . The initial position ( y E ) 0 of the ball is 12 m . The speed ( v B ) 0 of the ball is 20 m/s ( ↑ ) . Calculation: Show the v-t curve of the elevator and ball in Figure (1). Calculate the time ( t 2 ) using the relation: v E = ( v E ) 0 + a E t 2 Substitute 0 for ( v E ) 0 , 7.8 m/s for v E and 1.2 m/s 2 for a E . 7.8 = 0 + ( 1.2 × t ) t 2 = 7.8 1.2 t 2 = 6.5 sec Calculate the time ( t t o p ) at which the ball reaches the top of its trajectory: v B = ( v B ) 0 − g ( t t o p − t ) Here, t is time at the elevator begins to move and v B is final velocity of ball. Substitute 0 for v B , 20 m/s for ( v B ) 0 , 9.81 m/s 2 for g and 2 sec for t 2 . 0 = 20 − 9.81 ( t t o p − 2 ) t t o p = ( − 20 − 9.81 ) + 2 t t o p = 4.0387 sec Calculate the position ( y E ) of the elevator at time t top using the relation: ( y E ) t t o p = ( y E ) 0 + ( 1 2 a E t top 2 ) Substitute − 12 m for ( y E ) 0 , 1.2 m/s 2 for a E , and 4.0387 sec for t t o p . ( y E ) t t o p = − 12 + ( 1 2 × ( 1.2 ) × 4.0387 2 ) = − 12 + 9.787 = − 2.213 m Calculate the position ( y B ) of the ball when time ( t t o p ) is using the relation: ( y B ) t t o p = 1 2 ( v B ) 0 ( t top − t ) Substitute 20 m/s for ( v B ) 0 , 2 sec for t and 4.0387 sec for t t o p . ( y B ) t t o p = 1 2 × 20 × ( 4.0387 − 2 ) = 20.387 m Calculate the position ( y E ) t 2 of the elevator at time t 2 : ( y E ) t 2 = ( y E ) 0 + ( 1 2 v E t 2 ) Substitute − 12 m for ( y E ) 0 , 7

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ISBN: 9781260663778

Author: BEER

Publisher: MCG

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Chapter 11.3, Problem 11.75P

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When the ball will hit (t) the elevator.

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When the ball will hit ( t ) the elevator.

Solution Summary: The author shows the v-t curve of the elevator and ball in Figure (1).

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