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.

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 $u_y = u \sin \theta$.

Assuming no air resistance, the horizontal acceleration is $a_x = 0$.
The horizontal displacement $x$ after time $t$ is given by:

The acceleration due to gravity acts downwards, so $a_y = -g$.
The vertical displacement $y$ after time $t$ is given by:

From equation (1), solve for $t$:

Substitute this expression for $t$ into equation (2):

Simplify the first term:

Simplify the second term:

Recall that $\frac{1}{\cos^2 \theta} = \sec^2 \theta$.
Therefore, the equation for the trajectory of the projectile is:

This matches the required equation.