a stone falls from the top of a cliff into the ocean. In the air it had an average speed of 16m/s.In the water it had an average speed of 3m/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?

Let t represent the time the stone fell in the air.

distance = speed * time

t*16 + (12-t)*3 = 127 . . . . . total length of travel
13t = 91
t = 7 . . . . . time in air

The stone fell 7 seconds in the air, then 5 seconds in the water.

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parmesanchilliwack parmesanchilliwack · Answer 2 · 2019-04-10T09:22:40+02:00

Answer:

The stone fell in air for 7 seconds and fell in the water for 5 seconds,

Step-by-step explanation:

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,

Given,

Its average speed in air = 16 m/s,

And, its average speed in water = 3 m/s,

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 by it = 16 x + 3(12-x)

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

⇒ 16 x + 3(12-x) = 127

⇒ 16x + 36 - 3x = 127

⇒ 13x + 36 = 127

⇒ 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