Question

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 the air and how long did it fall in the water?

Answer 1

The time taken by stone in air is 7 seconds while the time taken by water is 12-7 = 5 seconds.

Explanation:

Given that,

The average speed of the stone is air is 16 m/s. Its average speed in seabed is 3 m/s. The total distance from the top of the cliff to the seabed is 127 meters, and the stone's entire fall took 12 seconds.

Let x represents the time taken by stone in the air,

Here, the total time taken by it in both air and water = 12 seconds,

Thus, the time taken by it in water = (12 - x) seconds,

Since, the distance covered by it in air = Time taken by it in the air × its speed in air = 16 x

Also, the distance covered by it in water = Time taken by it in the water × its speed in water= 3(12-x)

Total distance covered will be :

$D=16 x + 3(12-x)$

D = 127 m (given)

$16 x + 3(12-x)=127\\\\16x+36-3x=127\\\\13x + 36 = 127\\\\13x = 91 \\\\x=7\ s$

So, the time taken by stone in air is 7 seconds while the time taken by water is 12-7 = 5 seconds.

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