Hello everyone! I need your help with the next sketch:
I need to find the angle between the red line and the horizontal ($\alpha$). I did try everything that i could imagine but even so i couldn't find it.
According to the book, the angle should be: $\alpha=20~degrees$
How can i calculate it?
Hi I like Serena ! I'm not sure at all, but i think in that case $tan2\theta=-\dfrac{x}{y}$ which means that $2\theta$ is a complementary angle of $\alpha$ ... if that's correct, then $2\theta=C+\alpha$. Obviously C has to be $\pi/2$ but as i said, I'm not sure at all...
Hello! I have this equation:
$tan2\theta \cdot tan \alpha =-1$
where $\alpha$ is known. I need to find $\theta$. The answer is:
$\theta = \dfrac{\pi}{4}+\dfrac{\alpha}{2}$
How is that possible?
Hello Bacterius! Thank you very much for help me, as you said, I've been mistaken in the "split up" step. After correcting that mistake I was able to get your solution "$\dfrac{c}{2}$" in the form $\dfrac{v_0}{2mgsin\alpha }$ to finally get the expected solution.
Again thank you a lot ! It...
Hello guys ! I need your help with the next problem:
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"Show that the equation:
$\dfrac{mv_0}{k}-\dfrac{m^2g}{k^2}sin\alpha \cdot ln\left[ 1+\dfrac{kv_0}{mgsin\alpha } \right]$
is equivalent to:
$\dfrac{{v_0}^2}{2gsin\alpha...
Well actually... yes, the orginal problem is to demostrate that the differential equation $xdy-ydx=tan^{-1}(y/x)dx$ can be solved by using the substitution $y=vx$ even for this non-homogeneus equation. So i proceed to solve:
$y=vx$
Then
$dy=vdx+xdv$
Substituing in the original differential...
Hello everyone! I need some help with the integral:
$\displaystyle \int \dfrac{1}{\tan^{-1}(x)}dx$
I don't know how to solve it... can you guys help me please?