Question

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

Time in the air t₁ = 7 s

Time in the sea t₂ = 5 s

Explanation:

We will use the equation: v = d/t , where:

v is the average speed,

d the distance traveled, and

t time doing the trip

From v = d/t ⇒ d = v*t

Therefore, if we call t₁ time in the air, and t₂ time in the sea, according to problem statement, we have:

t₁ + t₂ = 12 (1)

16*t₁ + 3*t₂ = 127 (2)

A two equation system, from equation (1) we get

t₂ = 12 - t₁

And by substitution in equaton (2)

16*t₁ + 3 * ( 12 - t₁ ) = 127

16*t₁ + 36 - 3*t₁ = 127

13*t₁ = 91

t₁ = 91 /13

t₁ = 7 s

And t₂ = 12 - 7

t₂ = 5 s