A particle of mass � is projected from the ground with speed � at an angle � above the horizontal. Air resistance is negligible and gravity is �. Derive expressions for the time of flight, maximum height, and horizontal range. Show that the range is maximum when �. Now suppose the particle is projected from a height � above the ground instead of from ground level. Derive a general expression for the time of flight. For the same initial speed �, find the angle � that maximizes the range when launched from height �.

Here are the derivations for projectile motion: Part 1: Particle projected from the ground Let the initial speed be v and the launch angle be above the horizontal. The acceleration due to gravity is g. The initial velocity components are: Horizontal: v_x = v Vertical: v_y = v Time of Flight (T) The particle starts at y=0 and returns to y=0. Using the vertical displacement equation y = v_y t - (1)/(2)gt^2: 0 = (v ) T - (1)/(2)g