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 . The initial horizontal velocity is u_x = u . The initial vertical velocity is u_y = u . 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 + (1)/(2) a_x t^2 x = (u ) t + (1)/(2) (0) t^2 x = (u ) t (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 + (1)/(2) a_y t^2 y = (u ) t - (1)/(2) g t^2 (2) Step 4: Eliminate time (t) from the equations. From equation (1), solve for t: t = (x)/(u ) Substitute this expression for t into equation (2): y = (u ) ((x)/(u )) - (1)/(2) g ((x)/(u ))^2 Step 5: Simplify the equation. Simplify the first term: y = x (u )/(u ) = x Simplify the second term: y = x - (1)/(2) g (x^2)/(u^2 ^2 ) Recall that (1)/(^2 ) = ^2 . Therefore, the equation for the trajectory of the projectile is: y = x - (g x^2)/(2 u^2) ^2 This matches the required equation. 3 done, 2 left today. You're making progress.