To determine The height above the top of the window from which the object was released. Explanation Take the upward direction to be positive. The motion of the object is shown in the following figure . In the figure, Motion 1 is the motion of the object after releasing until it reaches the top of the window and Motion 2 is the motion of the object as it passes the window. Consider Motion 2 first. Write the equation of motion. y = y 0 + v 0 t + 1 2 a t 2 Here, y is the final position, y 0 is the initial position, v 0 is the initial speed, t is the time taken for the motion and a is the acceleration. The acceleration of the object will be equal to the negative of the acceleration due to gravity. Take the initial position of the object to be y 0 = 0 . Substitute 0 for y 0 and − g for a in the above equation. y = 0 + v 0 t + 1 2 ( − g ) t 2 y = v 0 t − 1 2 g t 2 v 0 t = y + 1 2 g t 2 v 0 = y t + 1 2 g t (I) Consider Motion 1. Write the equation of motion. v 2 = v 0 2 + 2 a ( y − y 0 ) Here, v is the final speed. The initial speed of the object will be zero and the acceleration of the object will be equal to the negative of the acceleration due to gravity. Take the initial position of the object to be y 0 = 0 . Substitute 0 for v 0 , 0 for y 0 and g for a in the above equation

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Chapter 2, Problem 73E

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The height above the top of the window from which the object was released.

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Chapter 2, Problem 72E

Chapter 2, Problem 74E

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