Is the way of determining my angle of vector wrong or am I using the wrong formula for calculating magnitude of resultant? [closed]

Question

i) Two balls A and B are simultaneously projected from top of building at 10m/s upward and 20m/s downward. Find distance between them after 3s? (Answer: 90m)

There are two magnitude of resultant formula: $$ First: R = |\vec{A} + \vec{B}| = \sqrt{(A^2 + B^2 + 2ABCos\theta)} ;R = |\vec{A} - \vec{B}| = \sqrt{(A^2 + B^2 + 2A(-B)Cos\theta)} $$ $$ Second: R = |\vec{A} + \vec{B}| = \sqrt{(|A^2| + |B^2| + 2|A||B|Cos\theta)} ;R = |\vec{A} - \vec{B}| = \sqrt{(|A^2| + |B^2| + 2|A||(-B)|Cos\theta)} $$ I think the first formula is right, so I will be solving the questions using first formula. Now for angle, opposite parallel vector angle is 180 and same side parallel vector is 0 degree.

Initial solution for i):

A = 10m/s, B = 20m/s, t = 3s, angle = 180(180 degrees as they are opposite parallel vectors) but right answer either comes from angle = 0 degree or using different formula. Lets assume angle is 180 as they are opposite parallel vectors. So, answer by this method is wrong. So, $$ R = |\vec{A} + \vec{B}| = \sqrt{(A^2 + B^2 + 2ABCos\theta)} = \sqrt{(10^2+20^2+2*10*20*Cos180)} = \sqrt{(100+400+400*-1)} = \sqrt{100} = 10m/s $$ $$ R.V = \frac{d}{t} => d = RV*t = 10*3 = 30m $$

Potential correct solution for i):

A = 10m/s, B = 20m/s, t = 3s, angle = 0 degree So, $$ R = |\vec{A} + \vec{B}| = \sqrt{(A^2 + B^2 + 2ABCos\theta)} = \sqrt{(10^2+20^2+2*10*20*Cos0)} = \sqrt{(100+400+400*1)} = \sqrt{900} = 30m/s $$ $$ R.V = \frac{d}{t} => d = RV*t = 30*3 = 90m $$

Potential correct solution for i):

A = 10m/s, B = 20m/s, t = 3s, angle = 180 degree but using $$ R = |\vec{A} - \vec{B}| = \sqrt{(A^2 + B^2 + 2A(-B)Cos\theta)} $$ So, $$ R = |\vec{A} - \vec{B}| = \sqrt{(A^2 + B^2 - 2ABCos\theta)} = \sqrt{(10^2+20^2-2*10*20*Cos180)} = \sqrt{(100+400-400*-1)} = \sqrt{500+400} = 30m/s $$ $$ R.V = \frac{d}{t} => d = RV*t = 30*3 = 90m $$

This means either my formula is wrong or my assumed angle is wrong. I know that when two vectors are moving away from each other, in straight line their velocities are added (A+B). Since one ball is moving upward while other is using downward, using $$ R = |\vec{A} + \vec{B}| $$ makes more sense. Yet, answer comes wrong when I used this formula with 180 degrees. Any help on what to use or is my angle assumed is wrong or formula is wrong.

Answer 1

Your approach is way off, and you really don't need to use vectors for this.

First find the distance traveled by the first ball:

$$d_A(t) = a_At^2+v_At$$

then find the distance traveled by the second ball:

$$d_B(t) = a_Bt^2+v_Bt$$

You know that the two balls are pointed in opposite directions, so think about the signs of the accelerations $a$ and the velocities $v$ (i.e. does acceleration add to the velocity or work against it?).

Then once you get each distance, you can simply add them together.

What you were doing is trying to find some kind of "average or resultant velocity" for the two objects, which is not physically meaningful.

Also, in general rather than trying to find the correct formula for a physics problem, it's best to examine what is physically happening, especially by drawing diagrams, and determine the relationships between quantities (in this case, how does distance traveled related to starting velocity, acceleration, and time elapsed), until you find a way to compute what is needed from what is known. No formula will work if it is applied incorrectly.

Answer 2

The relative velocity is the difference between the two velocity vectors. This is alwayas so, by definition if you want. It does not matter which way are the directions of the two vectors. So, if you really want to use vectors, you just calculate the magnitude of the difference. The general formula for difference has a minus in front of the third term, no matter how the vectors are oriented. If the cosine happens to be negative, then the sign will change to positive (this is when the angle is larger than 90 degrees). $$|\vec{A}-\vec{B}|^2=A^2+B^2-2AB\cos(\theta) $$ For $\theta =180^o$ this becomes $$|\vec{A}-\vec{B}|^2=A^2+B^2+2AB=(A+B)^2 \ (*)$$ so, obviously $$|\vec{A}-\vec{B}|^2=A+B $$ And this is the magnitude of the relative velocity which you use in the next step.There is no need to plug the numbers in equation (*) but even if you do, you get the same result as if you just add the magnitudes, of course.