(a) Explanation Given information: The velocity ( v 1 ) of parachutist is in free fall is − 200 km/h . The parachute is open at an altitude ( x 1 ) of 600 m . From position ( x 2 ) 586 m to ( x 3 ) 30 m the parachute is descend at constant rate ( v 2 ) of − 50 km/h . Calculation: Convert the kilometer per hour to meter per sec. Consider the velocity ( v 1 ) : v 1 = 200 km/h × ( 1 , 000 m 1 km ) × ( 1 hr 3 , 600 sec ) = − 55.556 m/s Consider the velocity ( v 2 ) . v 1 = − 50 km/h × ( 1 , 000 m 1 km ) × ( 1 hr 3 , 600 sec ) = − 13.889 m/s Tabulated the velocity ( v ) corresponding to time ( t ) in Table 1: t (sec) v ( m / s ) x (m) 0 -55.556 600 t 1 -13.889 586 t 2 -13.889 30 t 3 -0 0 Plot the v - t curve of parachute as in Figure (1). Plot the x - t curve of parachute as in Figure (2). Calculate time ( t 1 ) using the relation: ( x 2 ) − ( x 1 ) = t 1 ( v 1 + v 2 2 ) Substitute 586 m for x 2 , 600 m for x 1 , − 55.556 m/s for v 1 , and − 13.889 m/s for v 2 . 586 − 600 = t 1 ( − 55 (b) To determine The initial ( − a ) deceleration.

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259977251

Author: BEER

Publisher: MCG

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Chapter 11.3, Problem 11.66P

(a)

To determine

The time (ttotal) required for the parachutist to land after opening his parachute.

(b)

To determine

The initial (−a) deceleration.

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Chapter 11.3, Problem 11.65P

Chapter 11.3, Problem 11.67P

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The time ( t total ) required for the parachutist to land after opening his parachute.

Solution Summary: The author calculates the time required for the parachutist to land after opening his parachute. The velocity is (v_1).