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USPshnik [31]
2 years ago
14

A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now

? 3)How far is the ship from its starting point? 4)On what bearing is it now from its oringinal position?​
Mathematics
1 answer:
algol [13]2 years ago
8 0

The answers to the bearing problems are listed below:

  1. How far North is the ship now ___________ 38.82 km
  2. How far East is the ship now ____________ 144.89 km
  3. How far is the ship from its starting point_____241.48 km
  4. On what bearing is it now from its original position____035°

<h3>Meaning of bearing.</h3>

Bearing can defined as branch of mathematics that describes the accurate location of an object at any point in time.

<h3>Analysis</h3>

The answers to the problem from 1-3 can be gotten by using sin and cos to find the missing sides.

The final value can be gotten using the cosine rule

4). bearing from original position = cos^{-1} (a^{2} + b^{2}﹣c^{2}) / 2ab = 35°

In conclusion, the answers for each is given in the list above.

Learn more about Bearing: brainly.com/question/24142612

#SPJ1

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4.75 seconds

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The ball first goes vertically upwards, gravity decelerates the body and it momentarily comes to rest at a point in its upward trajectory, which is the point of maximum height. From this point, to the ground, the ball behaves as a freely dropped body.

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a = - 32.2 (since it is deceleration)

we know that <em>v = u + at </em>

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Also , we have <em>s = ut + (1/2)at²</em>

Here, s is the maximum height from the point where ball is thrown

So,     s = 55(1.7) - (0.5)(32.2)(1.7)(1.7)

⇒ s = 140 feet

So at a height of 140 feet + 10 feet (initial height) = 150 feet, the ball acts as a freely dropped body.

Here,      u = 0

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⇒ t = 3.05 sec

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