Explanation Given information: A skydiver weigh w = 80 kg jumps out of an airplane. k = 12 k g / s with a closed parachute and k = 70 k g / s with an open parachute.His initial velocity is zero. Formula used: Terminal velocity is − m g k , where m is mass, w is weight and k is air resistance constant. The general solution of the equation of form v ' = − ( k m ) ( v + m g k ) is v ( t ) = − m g k + C e − ( k m ) t . Calculation: From the given information, we have m = 80 k g k = 12 k g / s ( c l o s e d ) k = 70 k g / s ( o p e n e d ) g = 9.8 m / s 2 v ( 0 ) = 0 Terminal velocity is − m g k = − ( 80 ) ( 9.8 ) 12 = − 65.33 m/s Free-fall with air residence: v ' = − ( k m ) ( v + m g k ) The general solution for the above equation is v ( t ) = − m g k + C e − ( k m ) t . v ( t ) = − 65.33 + C e − ( 70 80 ) t

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Chapter 10.2, Problem 15E

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To find:The skydiver velocity at t=25 s if the parachute opens after 20 s of free-fall.

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Chapter 10.2, Problem 14E

Chapter 10.2, Problem 16E

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The skydiver velocity at t = 25 s if the parachute opens after 20 s of free-fall.

Solution Summary: The author calculates the skydiver velocity at t=25 s if the parachute opens after 20s of free-fall.

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To find:The skydiver velocity at t=25 s if the parachute opens after 20 s of free-fall.

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Chapter 10 Solutions