{"id":2050,"date":"2021-06-24T18:26:53","date_gmt":"2021-06-24T12:56:53","guid":{"rendered":"https:\/\/acejee.com\/blog\/?p=2050"},"modified":"2024-09-30T08:11:09","modified_gmt":"2024-09-30T02:41:09","slug":"trajectory-of-a-projectile","status":"publish","type":"post","link":"https:\/\/acejee.com\/blog\/trajectory-of-a-projectile\/","title":{"rendered":"Trajectory of a projectile"},"content":{"rendered":"\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t
\n\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t
\n\t\t\t\t\t\t\t\t\tThe trajectory of a projectile is nothing but the equation of the parabolic path travelled by the projectile. In other words, this equation $y = \\tan \\theta x – \\cfrac{g}{2u^2 \\cos^2 \\theta} x^2$, tells us the height, or, $y-$ coordinate of the projectile when it has travelled a distance $x$ along the ground or vice versa.\n\nSo, let’s see how we derive this equation\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t
\n\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t
\n\t\t\t\t\t

Equation of trajectory of a projectile<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t
\n\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t
\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Trajectory\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t
\n\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t
\n\t\t\t\t\t\t\t\t\t

Recall, the equations of motion of the projectile for horizontal and vertical direction<\/p>\n

Horizontal Direction : $x = u_x t$ $=u \\cos \\theta t$<\/p>\n

Vertical Direction : $y = u_y t -$ $\\cfrac{1}{2} gt^2$ $=u \\sin \\theta -$ $\\cfrac{1}{2} gt^2$<\/p>\n

Since we want the equation of trajectory $y = f(x)$, let’s go ahead and eliminate time variable $t$ from the two equations which gives us $y = \\tan \\theta x – \\cfrac{g}{2u^2 \\cos^2 \\theta} x^2$<\/p>\n

Now, let’s say that you have been given the equation of trajectory of the projectile as $y =$ $ax -$ $bx^2$, can you determine angle of projection, maximum height and range<\/p>\n

Angle of Projection = $\\left.\\cfrac{dy}{dx}\\right|_{t=0}$<\/p>\n

Maximum Height: Find $x_{top}$ for which $\\cfrac{dy}{dx} =0$, then maximum height, $H=$ $ax_{top}$ $- bx_{top}^2$<\/p>\n

Range: Find $x$ for which $y=0$ (other than $x=0$) and that will be the range<\/p>\n

With that, now let’s find out the radius of curvature at different points along the trajectory.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

\n\t\t\t\t\t\t
\n\t\t\t\t\t
\n\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t
\n\t\t\t\t\t\t\t\t\t

Projectile Motion | Important Questions | JEE PYQs<\/p>