Question

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?

Answer 1

The stone fall in air for 7 seconds and fall in water for 5 seconds.

Step-by-step explanation:

Total distance to sea bed = 127 m

Let the distance traveled in air be s.

Distance traveled in water = 127 - s

Average velocity in air = 16 m/s

Average velocity in water = 3 m/s

Distance traveled in air = Average velocity in air x Time in air

s = 16 x t₁

$t_1=\frac{s}{16}$

Distance traveled in water = Average velocity in water x Time in water

127-s = 3 x t₂

$t_2=\frac{127-s}{3}$

We have total time is 12 seconds

That is

                t₁ + t₂ = 12

                $\frac{s}{16}+\frac{127-s}{3}=12\\\\3s+2032-16s=576\\\\13s=1456\\\\s=112m$

We have

            s = 16 x t₁

            112 = 16 x t₁

             t₁ = 7 s

           t₁ + t₂ = 12

            7 + t₂ = 12    

             t₂ = 5 s

The stone fall in air for 7 seconds and fall in water for 5 seconds.