Explanation Given: The height of the cliff is 200 m . The distance travelled by the rock in the last half second of its flight is 45 m . Formula used: The rock covers 45 m during the last half-second of its motion, the initial speed of the rock can be obtained from the second equation of motion by equating the distance covered with the acceleration and the time taken by the rock. Write the expression for the speed of the rock at the point where the last half-second starts. s 1 / 2 = v 1 t + 1 2 g t 2 ........(1) Here, s 1 / 2 is the distance covered in last second, v 1 is the speed of the rock at the beginning of last half-second, g is the acceleration due to gravity and t is the time. The velocity just before the last half-second is acquired by the rock is attained by traveling from its maximum height to just 45 m above the ground. Write the expression for the maximum height of the rock above the cliff. h max = v 0 2 2 g Here, h max is the maximum height of the rock above the cliff and v 0 is the initial speed of the rock. Write the expression for the distance covered by the rock to achieve the velocity before the last half-second. v 1 2 = v max 2 + 2 g ( h c + h max ) Substitute v 0 2 2 g for h max in above expression. v 1 2 = v max 2 + 2 g ( h c + v 0 2 2 g ) ........(2) Here, v max is the speed of the object at maximum height and h c is the height of the cliff covered by rock

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Chapter 2, Problem 75P

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The initial speed of the rock as it was throw from the 200 m high cliff.

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Chapter 2, Problem 74P

Chapter 2, Problem 76P

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The initial speed of the rock as it was throw from the 200 m high cliff.

Solution Summary: The author explains how the initial speed of the rock can be obtained from the second equation of motion by equating the distance covered with the acceleration and the time taken.

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The initial speed of the rock as it was throw from the 200 m high cliff.

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