A stone fell from the top of a cliff into the ocean. In the air, it had an average speed of 161616 / text{m/s}m/sstart text, m, slash, s, end text. In the water, it had an average speed of 333 / text{m/s}m/sstart text, m, slash, s, end text before hitting the seabed. The total distance from the top of the cliff to the seabed is 127127127 meters, and the stone's entire fall took 12 seconds. How long did the stone fall in the air and how long did it fall in the water?

Question

Libby Benson

Physics

4 November, 12:09

A stone fell from the top of a cliff into the ocean. In the air, it had an average speed of 161616 / text{m/s}m/sstart text, m, slash, s, end text. In the water, it had an average speed of 333 / text{m/s}m/sstart text, m, slash, s, end text before hitting the seabed. The total distance from the top of the cliff to the seabed is 127127127 meters, and the stone's entire fall took 12 seconds. How long did the stone fall in the air and how long did it fall in the water?

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Answers (1)

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Aldo Haynes · Answer 1

Initial velocity in air, Vo = 0 m/s

Final velocity in air, Vi = 16 m/s

Initial velocity in water, Vf = 3 m/s

Total distance, S = 127 m

Total time, T = 12 s

Using the equation of motion,

(V - U) t = s

S = s1 + s2

Let T = t1 + t2

127 = (16 * t1) + 3 * (12 - t1)

127 = 16t1 + 36 - 3t1

91 = 13t1

t1 = 91/13

= 7 seconds

Time taken in air, t1 = 7 seconds

t2 = 12 - 7

= 5 seconds.