A body is projected under gravity from a fixed point with a speed of u at an angle of elevation θ. Show that the trajectory of the projectile is y = x tan θ - gx²/2u² sec²θ.

To show the trajectory of a projectile, we will use the equations of motion for constant acceleration, considering the horizontal and vertical components of the motion.

Step 1: Resolve the initial velocity into horizontal and vertical components. Let the initial speed be $u$ and the angle of elevation be $\theta$. The initial horizontal velocity is $u_x = u \cos \theta$. The initial vertical velocity is $u_y = u \sin \theta$.

Step 2: Write the equation for horizontal displacement. Assuming no air resistance, the horizontal acceleration is $a_x = 0$. The horizontal displacement $x$ after time $t$ is given by: $x = u_x t + \frac{1}{2} a_x t^2$ $x = (u \cos \theta) t + \frac{1}{2} (0) t^2$ $x = (u \cos \theta) t \quad (1)$

Step 3: Write the equation for vertical displacement. The acceleration due to gravity acts downwards, so $a_y = -g$. The vertical displacement $y$ after time $t$ is given by: $y = u_y t + \frac{1}{2} a_y t^2$ $y = (u \sin \theta) t - \frac{1}{2} g t^2 \quad (2)$

Step 4: Eliminate time ($t$) from the equations. From equation (1), solve for $t$: $t = \frac{x}{u \cos \theta}$ Substitute this expression for $t$ into equation (2): $y = (u \sin \theta) \left(\frac{x}{u \cos \theta}\right) - \frac{1}{2} g \left(\frac{x}{u \cos \theta}\right)^2$

Step 5: Simplify the equation. Simplify the first term: $y = x \frac{u \sin \theta}{u \cos \theta} = x \tan \theta$ Simplify the second term: $y = x \tan \theta - \frac{1}{2} g \frac{x^2}{u^2 \cos^2 \theta}$ Recall that $\frac{1}{\cos^2 \theta} = \sec^2 \theta$. Therefore, the equation for the trajectory of the projectile is: $y = x \tan \theta - \frac{g x^2}{2 u^2} \sec^2 \theta$ This matches the required equation.

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