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

1 Answer

6 votes

Time taken by stone in air is 7 seconds and time taken by stone in water is 5 seconds

Solution:

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

Given that the stone's entire fall took 12 seconds

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

time taken by stone in the air = x

time taken by stone in water = 12 - x

In the air, it had an average speed of 16 m/s

average speed in air = 16 m/s

We know that,

distance = speed x time

distance covered by it in air =
16 * x = 16x

distance covered in air = 16x

It had an average speed of 3 m/s before hitting the seabed

average speed in water = 3 m/s

distance covered by it in water =
3 * (12 - x) = 36 - 3x

distance covered in water = 36 - 3x

Then,

Total distance covered = distance covered in air + distance covered in water

Total distance covered = 16x + 36 - 3x = 13x + 36

But, the total distance covered by it = 127 meters ( Given )

Therefore,

13x + 36 = 127

13x = 127 - 36

13x = 91

x = 7

Hence, the time taken by stone in air = x seconds = 7 seconds,

And, the time taken by it in water = 12 - x = 12 - 7 = 5 seconds

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