In the air, it had an average speed of 16 1616 m/s m/sstart text, m, slash, s, end text. In the water, it had an average speed of 3 33 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 127 127127 meters, and the stone's entire fall took 12 1212 seconds.

Question

Davon

Mathematics

8 May, 10:03

In the air, it had an average speed of 16 1616 m/s m/sstart text, m, slash, s, end text. In the water, it had an average speed of 3 33 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 127 127127 meters, and the stone's entire fall took 12 1212 seconds.

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Answers (1)

Know the Answer?

Azul Nelson · Answer 1

t_a = 7 s

t_w = 5 s

Step-by-step explanation:

Given:

- The average speed in air v_a = 16 m/s

- The average speed in water v_w = 3 m/s

- The total distance D = 127 m

- The total time T = 12 s

Find:

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

Solution:

- Lets time taken by stone to drop through air as t_a = t_a.

- The time taken in water would be t_w = (12 - t_a)

- The total distance covered can be calculated as:

D = v_a*t_a + v_w * (12 - t_a)

127 = 16*t_a + 3 * (12 - t_a)

Simplify and solve for t:

91 = 16*t_a - 3t_a

13*t_a = 91

t_a = 7 s

t_w = 5 s