# Help again

A man is running up an incline planr maaking an angle $$\theta$$ with horizontal with a speed u .Rain drops falling at an angle $$\alpha$$ with the vertical appear to the man as if they are falling at an angle of $$45^o$$ with the horizontal. The speed of rain drops is

Note by Kyle Finch
3 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Take relative velocity along x and y axes and equate magnitude

- 3 years, 3 months ago

While talking about the rain, you took its direction in the south west direction. What prompted you to do that?? Why didn't you take it in the south-east direction?? In that case the x-component of relative velocity will be

ucos(theta)-vsin(alpha) and not ucos(theta)+vsin(alpha) as in this case...

- 3 years, 3 months ago

Both are equivalent. Just interchange $$\alpha$$ in my equation with $$-\alpha$$ to get to the result that you have mentioned. This disparity arises since I am measuring the angle in the clockwise direction, while you are measuring it in the anticlockwise direction.

- 3 years, 3 months ago

Yea, but the final expressions for the velocity are completely different...Then how are the results equivalent??

- 3 years, 3 months ago

Yes i solved it . Any way thanx for helping .

- 3 years, 3 months ago

$$v=u \sin {(\pi/4 -\theta)} \csc {(\pi/4 -\alpha)}$$

- 3 years, 3 months ago

- 3 years, 3 months ago

Can u explain in brief

- 3 years, 3 months ago