Motion of a projectile — FAQ

It is not essential, and it's much safer to derive them from the basic SUVAT equations each time. Understanding the derivation (e.g., setting $v_y=0$ for max height, $s_y=0$ for time of flight) is the key skill being tested. Memorised formulae only work for level ground and can lead to errors if misapplied.

In this case, the final vertical displacement, $s_y$, is not zero. For example, if a particle is projected from the top of a 10m cliff and lands in the sea, its final vertical displacement is $s_y = -10 \, \text{m}$ (taking the launch point as the origin). You would use this value in the SUVAT equation $s_y = u_y t + \frac{1}{2}a_y t^2$ to find the time of flight.

The equation of the trajectory, which we derived as $y = x \tan \theta - \frac{gx^2}{2U^2 \cos^2 \theta}$, is a quadratic function of $x$ (in the form $y = Ax - Bx^2$). The graph of a quadratic function is a parabola.

If air resistance is considered, the problem becomes much more complex. Air resistance is a type of drag force that opposes motion. It is often modelled as being proportional to velocity or the square of velocity. This means the acceleration is no longer constant, and the SUVAT equations do not apply. You would need to use calculus (forming and solving differential equations) to analyse the motion. This is covered in more advanced topics within Mechanics.