undefined Solution To Determine: Height of the cliff and the distance from its base where the diver hit the water. Solution: The height of the cliff is 44 m and the diver hit the water at a distance of 7.5 m from the base of the cliff. Explanation: Given: The initial speed ( v o )of the diver is 2.5 m/s.The diver is thrown horizontally and hits water intime tof 3.0 s. Formula used: From kinematic equation: y = y o + v o y t + 1 2 g t 2 y = y o + v o sin θ t + 1 2 g t 2 ... ( 1 ) On the other hand, the horizontal distance ( x ) is defined as: x = v o x t x = v o cos θ t ... ( 2 ) Where: y is vertical position reached by the diver. y o is in the initial vertical position from where the diver was launched. v o y is the initial vertical speed of the diver. g is acceleration due to gravity on Earth. t is the time elapsed. v o is theinitial speed of the diver. θ is the diver’s launch angle. x is the horizontal distance. v o x is the initial horizontal speed of the diver. Calculations: Isolate y o from equation (1): y o = y − v o y sin θ t + 1 2 g t 2 ... ( 3 ) Substitute values in equation (3) y o = 0 − 2.5 sin 0 o ( 3 ) + 1 2 ( 9.8 ) ( 3.0 ) 2 ... ( 5 ) y o = 44 m Therefore, the height of the cliff is 44 m. On the other hand, in the case of horizontal distance x , substitute values in equation (2): x = 2.5 cos 0 o ( 3.0 ) = 7.5 m Therefore, the diver hits the water at a distance of 7.5 m from the base of the cliff. Conclusion: Thus,the height of the cliff is 44 m and the diver hits the water at a distance of 7.5 m from the base of the cliff.

7th Edition

ISBN: 9780321929013

Author: Douglas C. Giancoli

Publisher: PEARSON

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