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Physics help needed!

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.
Reply 1
Avatar for Actaeon
Actaeon
11
Try drawing a vector diagram - you can then use trig and Pythagoras to work out the resultant. :smile:
Reply 2
Avatar for Marry97
Marry97
OP
2
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.
Reply 3
Avatar for Reety
Reety
12
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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Reply 4
Avatar for Marry97
Marry97
OP
2
So which one is the base? 1.4 or 1.8?
Reply 5
Avatar for Actaeon
Actaeon
11
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⁡θ=oppositehypotenuse \sin \theta = \frac{opposite}{hypotenuse}

cos⁡θ=adjacenthypotenuse \cos \theta = \frac{adjacent}{hypotenuse}

tan⁡θ=oppositeadjacent \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 :smile:
Reply 6
Avatar for Phichi
Phichi
11
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⁡θ=oppositehypotenuse \sin \theta = \frac{opposite}{hypotenuse}

cos⁡θ=adjacenthypotenuse \cos \theta = \frac{adjacent}{hypotenuse}

tan⁡θ=oppositeadjacent \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 :smile:


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.
(edited 9 years ago)
Reply 7
Avatar for Actaeon
Actaeon
11
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 :smile:
Reply 8
Avatar for Marry97
Marry97
OP
2
Now I get it. Thank you :smile:

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