Projectiles with drag need velocity-dependent acceleration

Choose fixed horizontal x and upward y axes. Let linear drag oppose the velocity relative to still air, with the same coefficient in both directions. The horizontal equation has no driving force and gives exponential decay of horizontal velocity. The vertical equation has a constant downward gravitational drive plus linear resistance. Shift vertical velocity by the downward terminal velocity to turn that equation into the same exponential relaxation problem. Initial horizontal and vertical velocities remain independent data even though the two motions share one time coordinate and one eventual landing event.

Integrate each velocity using the stated launch position. The resulting vertical position contains a downward linear-in-time term and a transient exponential term; it is not quadratic for all times. At short times compared with the relaxation time, expanding the exponential recovers the familiar vacuum displacement to leading order as drag tends to zero. This is a valuable algebra check because large terms in the exact vertical expression cancel in that limit. Directly setting the relaxation time to infinity inside uncancelled pieces can hide the finite gravitational term rather than proving the limit correctly.

Specify the landing surface, then solve the vertical position equation for the first physically relevant time at which the projectile reaches it. If launch and landing heights coincide, time zero is an algebraic root but is not the later landing. An elevated horizontal launch has a positive landing time determined by gravity and vertical drag. Substitute that same time into horizontal position to find range. Comparing horizontal displacements at equal elapsed time answers a different question from comparing ranges at equal landing height, because drag also changes the flight duration.

For a falling projectile under this linear model, downward speed approaches a finite value while horizontal speed approaches zero. The horizontal displacement approaches an asymptote if the model is extended indefinitely. A real projectile usually lands first, and changing density, wind, or drag coefficient can invalidate the indefinite extrapolation. An exact formula is exact only within its model assumptions. Explain whether the air is at rest in the selected frame, whether buoyancy and lift are negligible, and why a linear resistance law is reasonable over the trajectory's full speed range.

With quadratic drag, the x force contains horizontal velocity multiplied by the total relative speed, and the y force contains vertical velocity multiplied by that same speed. Changing the launch's horizontal component therefore changes the subsequent vertical drag as well. You cannot independently reuse a one-dimensional quadratic falling solution for the vertical motion of an oblique projectile. The components form a coupled differential equation. Numerical integration is often appropriate, but its step size must resolve velocity changes, and the landing event should be interpolated or solved rather than assigned to the first time step below ground.

Before trusting numerical output, verify the initial acceleration from the force law, run a smaller time step, and compare with a case that has an analytic answer. Setting drag to zero should recover vacuum motion; replacing quadratic resistance by the linear law should match the formulas above. Monitor the sign of drag power relative to still air: it must remove mechanical energy while gravity transfers energy between kinetic and potential forms. These checks catch sign mistakes and discretization errors separately. They do not establish that the chosen drag law describes a spinning ball or a turbulent wake in a real experiment.