Question

A stone falls from the top of a cliff into the ocean. in the air, it had an average speed of 161616 \text{m/s}m/sm, slash, s. in the water, it had an average speed of 333 \text{m/s}m/sm, slash, s 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 air and how long did it fall in the water? the stone fell in the air for seconds and fell in the water for seconds.

Answer 1

The motion of the stone consists of two separate motions: in air, with average speed $v_1 = 16 m/s$, and in the water, with average speed $v_2 = 3 m/s$.
The total distance covered is 127 m, and it is the sum of the distances covered in air and in the water:
$127 m = S = S_1 + S_2$ (1)
and the total time taken is the sum of the time the stone travels in air + the time the stone travels in the water:
$12 s = t = t_1 + t_2$
Since $S=vt$, we can rewrite  (1) as
$127 = v_1 t_1 + v_2 t_2 = 16 t_1 + 3 t_2$

We have now a system of 2 equations with 2 unknown variables:
$\left \{ {{16t_1 + 3 t_2 = 127} \atop {t_1 + t_2 = 12}} \right.$
And if we solve it, we find
$t_1 = 7 s$
$t_2 = 5 s$
Which means that the stone travels for 7 seconds in air and for 5 seconds in the water.

Answer 2

Answer:

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

Explanation: