The trajectory of a projectile in a vertical plane is y = $\alpha$x $-$ $\beta$x2, where $\alpha$ and $\beta$ are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $\theta$ and the maximum height attained H are respectively given by :
Solution
y = $\alpha$x $-$ $\beta$x<sup>2</sup><br><br>comparing with trajectory equation<br><br>$y = x\tan \theta - {1 \over 2}{{g{x^2}} \over {{u^2}{{\cos }^2}\theta }}$<br><br>$\tan \theta = \alpha \Rightarrow \theta = {\tan ^{ - 1}}\alpha$<br><br>$\beta = {1 \over 2}{g \over {{u^2}{{\cos }^2}\theta }}$<br><br>$\Rightarrow$ ${u^2} = {g \over {2\beta {{\cos }^2}\theta }}$<br><br>Maximum height H :<br><br>$$H = {{{u^2}{{\sin }^2}\theta } \over {2g}} = {g \over {2\beta {{\cos }^2}\theta }}{{{{\sin }^2}\theta } \over {2g}}$$<br><br>$\Rightarrow$ $H = {{{{\tan }^2}\theta } \over {4\beta }} = {{{\alpha ^2}} \over {4\beta }}$
About this question
Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line
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