Question

THE velocity of a particle when its greatest height is√(2/5) of its velocity at half its greatest height.Find angle of projection of body.

Answer 1

read the question again plzz

Answer 2

\[\sqrt{2/5}\] or \[\sqrt{2}/5\] ???

Answer 3

[\sqrt{2/5}\] o

Answer 4

0 - start position 1-half position 2- full position
\[v_{0x}=v_{1x}=v_{2x}=v_2\]
\[\frac{v_2}{v_1}=\sqrt{\frac{2}{5}}\quad\Rightarrow\quad\frac{v^2_2}{v^2_1}=\frac{2}{5}\quad,\quad\frac{v^2_2}{v^2_{1x}+v^2_{1y}}=\frac{v^2_2}{v^2_2+v^2_{1y}}=\frac{2}{5}\]
\[v^2_{1y}=\frac{3}{2}v^2_{2}\quad\Rightarrow\quad v_{1y}=\sqrt{\frac{3}{2}}v_{2}\]
\[h_\max=\frac{v^2_{0y}}{2g}\quad,\quad v^2_{1y}=v^2_{0y}-2g\frac{h_\max}{2}=v^2_{0y}-g\frac{v^2_{0y}}{2g}=\frac{v^2_{0y}}{2}\]
\[\Rightarrow\quad v^2_{0y}=2v^2_{1y}\quad\Rightarrow\quad v_{0y}=\sqrt{2}v_{1y}=\sqrt{2}\sqrt{\frac{3}{2}}v_{2}=\sqrt{3}v_2\]
\[\tan\alpha=\frac{v_{0y}}{v_{0x}}=\frac{\sqrt{3}v_2}{v_{2}}=\sqrt{3}\quad\Rightarrow\quad\alpha=\frac{\pi}{3}=60^{o}\]

Answer 5

wow

Answer 6

thx 2 more questions

Answer 7

k?

Answer 8

what questions?

Answer 9

1.If at any instant the velocity of a projectile be u and its direction of motion theta with horizontal then show that it will be moving at right angles to this direction after a time (u/g)*cosec theta

Answer 10

\[t_{top}=\frac{u\sin{\theta}}{g}<\frac{u}{g\sin\theta}\quad,\quad\Delta t=\frac{u}{g\sin\theta}-\frac{u\sin\theta}{g}=\frac{u}{g}\left(\frac{1}{\sin\theta}-\sin\theta\right)\]
\[=\frac{u}{g}\frac{1-\sin^2\theta}{\sin\theta}=\frac{u}{g}\frac{\cos^2\theta}{\sin\theta}\]
\[\vec{u}=<u\cos\theta,u\sin\theta>\quad,\quad\vec{v}=<u\cos\theta,-g\Delta t>=<u\cos\theta,-u\frac{\cos^2\theta}{\sin\theta}>\]
\[\vec{u}\cdot\vec{v}=<u\cos\theta,u\sin\theta>\cdot<u\cos\theta,-u\frac{\cos^2\theta}{\sin\theta}>=\]\[=u^2\cos^2\theta-u^2\cos^2\theta=0\quad\Rightarrow\quad\vec{u}\perp\vec{v}\]

Answer 11

\[\frac{u}{g\sin\theta}\leq\frac{2u\sin\theta}{g}\]
\[\frac{1}{\sin\theta}\leq 2\sin{\theta}\quad,\quad 1\leq 2\sin^2{\theta}\quad,\quad \frac{1}{\sqrt{2}}\leq\sin\theta\quad,\quad\frac{\pi}{4}\leq\theta<\frac{\pi}{2}\]

Answer 12

second question?