Marry97
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#1
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A rain drop falls at a constant vertical velocity of 1.8m/s in still air and a horizontal wind is blowing. If the velocity of the wind is 1.4m/s, determine the magnitude and direction of the resultant velocity of the rain drop.
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Actaeon
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Try drawing a vector diagram - you can then use trig and Pythagoras to work out the resultant.
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Marry97
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I found the magnitude of the resultant using Pythagoras but I can't find the angle using trigonometry because I dot know what to the base and perpendicular are.
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Reety
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(Original post by Marry97)
I found the magnitude of the resultant using Pythagoras but I can't find the angle using trigonometry because I dot know what to the base and perpendicular are.
It's the same values you used at the start.


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Marry97
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So which one is the base? 1.4 or 1.8?
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Actaeon
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The sides you use in trig formulae depend on the angle you choose in the first place. There are three sides; the side opposite your angle, the side adjacent to your angle, and the hypotenuse.

You choose your angle, and can use the mnemonic SOH CAH TOA to remember the order -

 \sin \theta = \frac{opposite}{hypotenuse}

 \cos \theta = \frac{adjacent}{hypotenuse}

 \tan \theta = \frac{opposite}{adjacent}

So in this case, choose your angle, work out which side is opposite to it and adjacent (next) to it, and plug in your values
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Phichi
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(Original post by Actaeon)
The sides you use in trig formulae depend on the angle you choose in the first place. There are three sides; the side opposite your angle, the side adjacent to your angle, and the hypotenuse.

You choose your angle, and can use the mnemonic SOH CAH TOA to remember the order -

 \sin \theta = \frac{opposite}{hypotenuse}

 \cos \theta = \frac{adjacent}{hypotenuse}

 \tan \theta = \frac{opposite}{adjacent}

So in this case, choose your angle, work out which side is opposite to it and adjacent (next) to it, and plug in your values
I think he's confused at which value is which, not what the sides are called and how to apply trig.

@OP, you have 1.8m/s going vertically down, and 1.4m/s going horizontally, you can draw a right angled triangle and find the angle from the vertical.
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Actaeon
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(Original post by Phichi)
I think he's confused at which value is which, not what the sides are called and how to apply trig.

@OP, you have 1.8m/s going vertically down, and 1.4m/s going horizontally, you can draw a right angled triangle and find the angle from the vertical.
Ah right! Though from the diagram it should be clear which value is which, and given the talk about 'base' and 'perpendicular', I wasn't sure whether or not they were clear on using trig
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Marry97
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#9
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Now I get it. Thank you
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