Question

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 121212 seconds. How long did the stone fall in the air and how long did it fall in the water?

Answer 1

Answer:

The amount of time the stone fall in the air, t₁, is 7 seconds and

The amount of time the stone fall in the water, t₂, is 5 seconds

Explanation:

Here, we have the average speed of the stone in air = 16 m/s

The average speed of the stone in water = 3 m/s

Total distance from top of cliff to sea bed = 127 meters

Total time of fall = 12 seconds

Therefore

Let t₁ = time of the stone in air and

t₂ = time of the stone in water

Therefore, t₁ + t₂ = 12 s........(1)

16·t₁ + 3·t₂ = 127............(2)

From (1) t₁ = 12 - t₂

16×(12 - t₂) + 3×t₂ = 127

From which 13·t₂ = 65

and t₂ = 5 seconds

∴ t₁ = 12 - t₂ = 12 - 5 = 7 seconds

The amount of time the stone fall in the air is t₁ = 7 seconds and

The amount of time the stone fall in the water is t₂ = 5 seconds.