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

Answer:

a) the angle of ascent is 8.2°

b) the horizontal distance traveled is 4375 m

Step-by-step 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.