Explanation Given information: Mass of skydiver = 80 kg Formula used: The force of gravity apply to any falling object of mass in is equal to mass m is equal to − m g where g is the acceleration due to gravity applied by earth to any mass m . Differential Equation for velocity is given as; u ' = − k m ( u + m g k ) Where u is the initial velocity and k be and constant The solution of the differentiation equation will be; u ' = − k m ( u + m g k ) u ( t ) = − m g k + C e − ( k m ) t Calculation: As the skydiver jumps with zero initial velocity. Therefore, u ( 0 ) = 0 put in the above equation to find the value of C . u ( t ) = − m g k + C e − ( k m ) t 0 = − m g k + C e − ( k m ) 0 = − m g k + C Hence use k = 12 kg/sec and g = 9.8 m/sec 2 as initially the parachute was closed, to obtain C = m g k = 80 ( 9.8 ) 12 = 65.33 In order to find the velocity at t = 25 sec . First find the velocity of the skydiver after t = 20 sec , that is during free fall under the influence of air resistance and use k = 12 kg/sec

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ISBN: 9781319545956

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 9.2, Problem 15E

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The skydiver’s velocity at t=25 s if the parachute opens after 20 s of free fall.

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Chapter 9.2, Problem 14E

Chapter 9.2, Problem 16E

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The skydiver’s velocity at t = 25 s if the parachute opens after 20 s of free fall.

Solution Summary: The author calculates the skydiver's velocity at t=25s if the parachute opens after free fall.

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The skydiver’s velocity at t=25 s if the parachute opens after 20 s of free fall.

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The skydiver’s velocity at t = 25 s if the parachute opens after 20 ​ s of free fall.

Solution Summary: The author calculates the skydiver's velocity at t=25s if the parachute opens after free fall.

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The skydiver’s velocity at t=25 s if the parachute opens after 20​ s of free fall.

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Chapter 9 Solutions

The skydiver’s velocity at t = 25 s if the parachute opens after 20 s of free fall.

The skydiver’s velocity at t=25 s if the parachute opens after 20 s of free fall.