Explanation Draw the figure as per given situation as. When the bomb has fallen the vertical distance of 2.15 km , it will also travel some horizontal distance. Refer to the above figure (b), use the Pythagoras theorem for horizontal distance. Write the expression for horizontal distance as. x = ( A C ) 2 − ( A B ) 2 (I) Here, x is horizontal distance, A C is displacement and A B is vertical distance. Write the general expression for horizontal displacement as. x = v i cos θ t (II) Here, x is horizontal displacement, v i cos θ is initial velocity along horizontal and t is time. Rearrange above equation (II) in terms of v i . v i = x cos θ ( t ) Rearrange the equation (II) in terms of t . t = x v i cos θ Write the expression for vertical displacement as y = v i sin θ t − 1 2 g t 2 (III) Here, y is vertical displacement, v i sin θ is initial velocity along vertical and g is acceleration due to gravity. Substitute x cos θ ( t ) for v i and x v i cos θ for t in equation (III). y = x cos θ ( t ) sin θ ( t ) − 1 2 g ( x v i cos θ ) 2 Rearrange and simplify the above equation as. y = x tan θ − g x 2 2 v i 2 cos 2 θ (IV) Substitute tan 2 θ + 1 for 1 cos 2 θ in equation (III). y = x tan θ − g x 2 ( tan 2 θ + 1 ) 2 v i 2 Rearrange the above expression for quadratic equation in terms of tan θ . g x 2 2 v i 2 tan 2 θ − x tan θ + g x 2 2 v i 2 + y = 0 (V) Conclusion: Substitute 3.25 km for A C and 2.15 km for A B in equation (I). x = ( 3

Physics for Scientists and Engineers with Modern Physics, Technology Update

9th Edition

ISBN: 9781305401969

Author: SERWAY, Raymond A.; Jewett, John W.

Publisher: Cengage Learning

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Chapter 4, Problem 85CP

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The angle θ.

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Chapter 4, Problem 84CP

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Chapter 4 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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Chapter 4 Solutions