JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 12)
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 :
$${\tan ^{ - 1}}\alpha ,{{{\alpha ^2}} \over {4\beta }}$$
$${\tan ^{ - 1}}\alpha ,{{4{\alpha ^2}} \over \beta }$$
$${\tan ^{ - 1}}\left( {{\beta \over \alpha }} \right),{{{\alpha ^2}} \over \beta }$$
$${\tan ^{ - 1}}\beta ,{{{\alpha ^2}} \over {2\beta }}$$
Explanation
y = $$\alpha$$x $$-$$ $$\beta$$x2
comparing with trajectory equation
$$y = x\tan \theta - {1 \over 2}{{g{x^2}} \over {{u^2}{{\cos }^2}\theta }}$$
$$\tan \theta = \alpha \Rightarrow \theta = {\tan ^{ - 1}}\alpha $$
$$\beta = {1 \over 2}{g \over {{u^2}{{\cos }^2}\theta }}$$
$$ \Rightarrow $$ $${u^2} = {g \over {2\beta {{\cos }^2}\theta }}$$
Maximum height H :
$$H = {{{u^2}{{\sin }^2}\theta } \over {2g}} = {g \over {2\beta {{\cos }^2}\theta }}{{{{\sin }^2}\theta } \over {2g}}$$
$$ \Rightarrow $$ $$H = {{{{\tan }^2}\theta } \over {4\beta }} = {{{\alpha ^2}} \over {4\beta }}$$
comparing with trajectory equation
$$y = x\tan \theta - {1 \over 2}{{g{x^2}} \over {{u^2}{{\cos }^2}\theta }}$$
$$\tan \theta = \alpha \Rightarrow \theta = {\tan ^{ - 1}}\alpha $$
$$\beta = {1 \over 2}{g \over {{u^2}{{\cos }^2}\theta }}$$
$$ \Rightarrow $$ $${u^2} = {g \over {2\beta {{\cos }^2}\theta }}$$
Maximum height H :
$$H = {{{u^2}{{\sin }^2}\theta } \over {2g}} = {g \over {2\beta {{\cos }^2}\theta }}{{{{\sin }^2}\theta } \over {2g}}$$
$$ \Rightarrow $$ $$H = {{{{\tan }^2}\theta } \over {4\beta }} = {{{\alpha ^2}} \over {4\beta }}$$
Comments (0)
