Confusion on resolving momentum

6

OK, so when solving this I did the parts of resolving the horizontal and vertical vectors correctly. I correctly calculated Horizontal velocity to be 1.32 m/s and correctly calculated vertical velocity to be 1.33 m/s. OK perfectly understood all this and makes sense.

However, what I don't understand is that the question is left to this, in calculation they consider vertical component to be the answer. This does not make any sense whatsoever as they did not ask me for the vertical component they asked me for "v".

Common sense and my understanding of physics tells me that v should be
found using Pythagoras theorem wherein v = √(1.31)^2 + (1.33)^2 which is equals to roughly 1.868. But such is not the case in the answer. I don't understand - explain please.

Reply 1

Eimmanuel

Study Forum Helper

15

Original post

by tigrispanthera

OK, so when solving this I did the parts of resolving the horizontal and vertical vectors correctly. I correctly calculated Horizontal velocity to be 1.32 m/s and correctly calculated vertical velocity to be 1.33 m/s. OK perfectly understood all this and makes sense.

However, what I don't understand is that the question is left to this, in calculation they consider vertical component to be the answer. This does not make any sense whatsoever as they did not ask me for the vertical component they asked me for "v".

Common sense and my understanding of physics tells me that v should be
found using Pythagoras theorem wherein v = √(1.31)^2 + (1.33)^2 which is equals to roughly 1.868. But such is not the case in the answer. I don't understand - explain please.

It seems that you have a mental block.

Indeed, the question asked for the speed of the stone, v.

Note that the vertical component of the lower stone’s velocity is
$– v \sin(30^{\circ}) = - 0.5v$

NOT v.

You may want to read the quote a few times and slowly. :smile:

The correct application of Pythagoras' theorem (or if you like correct common sense) to find the magnitude of the velocity is

$\sqrt{(– v \sin(30^{\circ}))^2 + (v \cos(30^{\circ}))^2 }$