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A stone fell from the top of a cliff into the ocean. In the air, it had an average speed of 16 m/s. In the water, it had an average speed of 3 m/s before hitting the seabed. The total distance from the top of the cliff to the seabed is 127 meters, and the stone's entire fall took 12 seconds.How long did the stone fall in air and how long did it fall in the water?

Respuesta :

Answer:

Stone falls in air for t = 7 s

and stone falls in water for t = 5 s

Explanation:

As we know that average speed of the stone in air is given as

[tex]v_{air} = 16 m/s[/tex]

average speed of stone in water is given as

[tex]v_{water} = 3 m/s[/tex]

total time of the motion of stone in water and air is given as

[tex]T = 12 s[/tex]

total distance from the cliff to the seabed is given as

[tex]d = 127 m[/tex]

let the stone moves in air for time "t"

and stone mover in water for total time "12 - t"

now we have

[tex]d = v_1 t + v_2 (12 - t)[/tex]

[tex]127 = 16 t + 3(12 - t)[/tex]

[tex]127 = 13 t + 36[/tex]

[tex]t = 7 seconds[/tex]

so stone falls in air for t = 7 s

and stone falls in water for t = 5 s

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