Question

physics +math doubts

Answer 1

3 log3 13 : simplify

Answer 2

yey

Answer 3

shall i post it?

Answer 4

chech??

Answer 5

A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the othr end of the base .If alpha nd beta are base angles ,and theta the angle of projection ,prove that tan theta=tan alpha +tan beta

Answer 6

\[d_\max=h\left(\frac{1}{\tan\alpha}+\frac{1}{\tan\beta}\right)=\frac{v_0^2}{g}\sin{2\theta}\]\[y=v_0\sin\theta\cdot t-\frac{1}{2}gt^2=v_0\sin\theta\cdot \frac{x}{v_0\cos\theta}-\frac{1}{2}g\left(\frac{x}{v_0\cos\theta}\right)^2=\]\[=x\tan\theta-\frac{g}{2v_0^2\cos^2\theta}x^2\]\[h=y\left(x=\frac{h}{\tan\alpha}\right)=\frac{h}{\tan\alpha}\tan\theta-\frac{g}{2v_0^2\cos^2\theta}\left(\frac{h}{\tan\alpha}\right)^2\]\[1=\frac{\tan\theta}{\tan\alpha}-\frac{g}{2v_0^2\cos^2\theta\tan^2\alpha}h=\frac{\tan\theta}{\tan\alpha}-\frac{g}{2v_0^2\cos^2\theta\tan^2\alpha}\cdot\frac{v_0^2}{g}\sin{2\theta}\frac{\tan\alpha\cdot\tan\beta}{\tan\alpha+\tan\beta}\]\[1=\frac{\tan\theta}{\tan\alpha}-\frac{\sin{2\theta}\cdot\tan\alpha\cdot\tan\beta}{2\cos^2\theta\cdot\tan^2\alpha\cdot(\tan\alpha+\tan\beta)}=\frac{\tan\theta}{\tan\alpha}-\frac{\sin{\theta}\cdot\tan\beta}{\cos\theta\cdot\tan\alpha\cdot(\tan\alpha+\tan\beta)}\]\[1=\frac{\tan\theta}{\tan\alpha}-\frac{\tan{\theta}\cdot\tan\beta}{\tan\alpha\cdot(\tan\alpha+\tan\beta)}=\frac{\tan\theta}{\tan\alpha}\left(1-\frac{\tan\beta}{\tan\alpha+\tan\beta}\right)=\frac{\tan\theta}{\tan\alpha+\tan\beta}\]\[\Rightarrow\quad\tan\theta=\tan\alpha+\tan\beta\]