Question

Whale sharks swim forward while ascending or descending. They swim along a straight-line path at a shallow angle as they move from the surface to deep water or from the depths to the surface. In one recorded dive, a shark started 50 m below the surface and swam at 0.85 m/s along a path tipped at a 13 ∘ angle above the horizontal until reaching the surface.

Answer 1

You did not include the quesetion, but I can help you to understand the problem and how to find the relevant information.

1) The angle of 13° with which the shark ascends meets this:

Vertical ascending velocity = 0.85m/s * sin(13°)

Horizontal velocity = 0.85m/s * cos(13°)

2) The length swan by the shark ascending meets this

Vertical ascending length = 50 m

Horizontal length, y:

$\frac{y}{50} = \frac{0.85sin(13)}{0.85cos(13)}$

From that y = 50 * tan(13°)

=> y = 11.54 m.

3) Conclusions:

1) The shark run 50 m vertically upward and 11.54 m horizontally.

2) The length of the path run by the shark may be calculated using Pythagoras' theorem:

hypotenuse^2 = (50m)^2 + (11.54m)^2 = 2633.25m^2

hypotenuse = 51.35m

So, the shark swan 51.35 m to reach the surface.