Explanation Given: Time difference between drop and throw of two stone is 1.6 s . Initial speed of throw for second stone is 32 m / s . Formula used: The distance covered in both drop and throw is same. Only difference is initial velocity and time of throw. Write the expression of displacement in terms of initial velocity, time, acceleration due to gravity s = v 0 t + 1 2 g t 2 ..............(1) Here, s is displacement covered, v 0 is initial velocity and t is the time. Write the expression for displacement for dropped stone. s 1 = 1 2 g t 1 2 .............. (2) Here, s 1 is displacement covered by dropped stone. Write the expression for displacement of thrown stone. s 2 = v 01 t 2 + 1 2 g t 2 2 .............. (3) Here, v 01 is initial velocity of thrown stone and t 2 is time taken to reach the lake. Displacement covered by both stones are same. Equate equations (1) and (2). 1 2 g t 1 2 = v 01 t 2 + 1 2 g t 2 2

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Author: Tipler

Publisher: MAC HIGHER

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Chapter 2, Problem 87P

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Height of the cliff.

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Chapter 2, Problem 86P

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Height of the cliff.

Solution Summary: The author calculates the distance covered in both drop and throw, based on initial velocity, time, and acceleration due to gravity.

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