Astone falls from the top of a cliff into the ocean. in the air, it had an average speed of 16 m/s slash, s. in the water, it had an average speed of 3 m/s , slash, 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?

Total distance = sum of velocity * time for both legs of the journey; let the time taken in air be t hence time taken in water is 12-t 127 = 16*t + 3(12-t) 127 = 16t + 36 - 3t 127 - 36 = 16t - 3t 91 = 13t t = 91/13 = 7 seconds  Hence time taken in air = 7 seconds time taken in water is 12 - 7 = 5 seconds checking; 16*7 + 3*5 = 127

The proof is mentioned below.

Step-by-step explanation:

Here, Δ ABC is isosceles triangle.

Therefore, AB = BC

Prove: Δ ABO ≅ Δ ACO

In Δ ABO and Δ ACO,

∠ BAO ≅ ∠ CAO  ( AO bisects ∠ BAC )

∠ AOB ≅ ∠ AOC ( AO is perpendicular to BC )

BO ≅ OC ( O is the mid point of BC)

Thus, By ASA postulate of congruence,

Δ ABO ≅ Δ ACO

Therefore, By CPCTC,

∠B ≅ ∠ C

Where ∠ B and ∠ C are the base angles of Δ ABC.