Explanation Given information: For a uniformly accelerated motion, x = x 0 + v 0 t + 1 2 a t 2 In the above equation, x - End distance x 0 - Start distance v 0 - Initial velocity t - Elapsed time a -Acceleration Calculation: For horizontal motion, x 0 = v 0 cos α t Therefore, v 0 = x 0 cos α t For vertical motion, y = y 0 + v 0 t + 1 2 a t 2 y = 0 + v 0 sin α t − 1 2 g t 2 Substitute for v 0 y = ( x 0 cos α t ) sin α t − 1 2 g t 2 y = x 0 tan α − 1 2 g t 2 Therefore, t 2 = 2 ( x 0 tan α − y ) g At the landing point, y = 0 Therefore, t = 2 v 0 sin α g x = 2 v 0 2 sin α cos α g Now, for α = 31 ° To clear the tree A, x A = 30 m y A = 12 m Therefore, t A 2 = 2 ( x 0 tan α − y ) g = 2 ( 30 tan 31 ° − 12 ) 9 To determine (b) v 0 , d if α = 27 °

Vector Mechanics for Engineers: Dynamics

11th Edition

ISBN: 9780077687342

Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 11, Problem 11.188RP

To determine

(a)

The velocity v0,d if α=31°

To determine

(b)

v0,d if α=27°

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Chapter 11, Problem 11.187RP

Chapter 11, Problem 11.189RP

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