Question

How do you calculate the resultant of the following vectors 4.0m/s and 10m/s with an angle of 90degrees between them?

Answer 1

Suppose `vec a=4 m/s` and `vec b=10 m/s`

So,resultant `vec c = vec a +vec b`

So, `|vec c| = sqrt(a^2+b^2+2ab cos theta)` (where, `theta` is the angle between `vec a` and `vec b`)

Thus from the above formula we get, magnitude of the resultant vector is `sqrt(4^2+10^2+2*410 cos 90)=10.77 m/s`

Making an angle of `tan ^-1(4/10)=21.80` degrees w.r.t `vec b`