Question

Un corp este aruncat cu viteza inițială v0=20m/s sub un unghi α0=45° fată de orizontala.Dupa cat timp si la ce înălțime vectorul viteza formează un unghi α=30° cu orizontala?

Answer 1

$\displaystyle \text{Se da:}\\ \\v_0=20\frac ms\\ \\\alpha_0=45^\circ \\ \\\alpha=30^\circ\\ \\t=?s\\ \\h=?m\\ \\ \\\text{Formule:}\\ \\\text{Folosind unghiul, putem afla proiectia vitezei pe orizontala:}\\\\v_x=v_0\times\cos{\alpha_0}\\ \\\text{Pentru orice unghi alfa, stim formula: }\\\\\tan{\alpha}=\frac{v_y}{v_x}\\\\\text{E important sa stim ca viteza pe orizontala va fi mereu aceeasi, astfel:}\\\\v_{y}=v_x\times\tan{\alpha}\\\\v_y=v_0\times\cos\alpha_0\times\tan\alpha$

$\displaystyle\text{La fel folosind unghiul, putem afla viteza initiala pe verticala:}\\\\v_{y_0}=v_0\times\sin\alpha_0\\\\\text{Socotind ca miscarea este una accelerata incetinita:}\\\\v_y=v_{y_0}-g\times t\\\\g\times t=v_0\times\sin\alpha_0-v_0\times\cos\alpha_0\times\tan\alpha\\ \\\\\boxed{t=\frac{v_0\times(\sin\alpha_0-\cos\alpha_0\times\tan\alpha)}{g}}\\ \\ \\\\\text{La fel:}\\\\v_y^2=v_{y_0}^2-2\times g\times h$

$\displaystyle \boxed{h=\frac{v_0^2\times(\sin^2\alpha_0-\cos^2\alpha_0\times\tan^2\alpha)}{2\times g}}\\ \\\\\text{Calcule:}\\ \\t=\frac{20\times(\sin45-\cos45\times\tan30)}{9,8}\approx0,61s\\\\h=\frac{20^2\times(\sin^245-\cos^245\times\tan^230)}{2\times 9,8}\approx 6,8m$