Question

A submarine is moving parallel to the surface of the ocean at a depth of 626 m. It begins a

constant ascent so that it will reach the surface after travelling a distance of 4420 m.
a) What angle of ascent, to the nearest tenth of a degree, did the submarine make? (3
marks)
b) How far did the submarine travel horizontally, to the nearest metre, during its ascent to
the surface? (3 marks)

Answer 1

a) the angle of ascent is 8.2°

b) the horizontal distance traveled is 4375 m

Explanation:

depth of ocean = 626 m

distance traveled in the ascent = 4420 m

This is an angle of elevation problem with

opposite side to the angle = 626 m

hypotenuse side = 4420 m

a) angle of ascent ∅ is gotten from

sin ∅ = opp/hyp = 626/4420

sin ∅ = 0.142

∅ =
`sin^(-1)` 0.142

∅ = 8.2° this is the angle of ascent of the submarine.

b) The horizontal distance traveled will be gotten from Pythagoras theorem

`hyp^(2)` =
`opp^(2)` +
`adj^(2)`

The horizontal distance traveled will be the adjacent side of the right angle triangle formed by these distances

`4420^(2)` =
`626^(2)` +
`adj^(2)`

adj =
`\sqrt{4420^(2)-626^(2) }`

adj = 4375 m this is the horizontal distance traveled.