Projectile Motion

With air resistance neglected, the only force on a projectile in flight is its weight, which acts vertically downwards. So there is no horizontal force, and by Newton’s first law the horizontal velocity is $\text{constant}$.

Vertically, the weight produces a constant acceleration $g$ downwards, so the vertical motion is standard uniformly-accelerated suvat.

The two directions share nothing except the time of flight. That is the hinge of every problem: find $t$ from one direction, then use it in the other.

When something is launched horizontally from a height, the initial vertical velocity is zero. That simplifies the vertical equation to $s_y=\frac{1}{2}g{t}^2$, so the time of flight depends $\text{only}$ on the drop height — the launch speed is irrelevant to how long it takes to fall.

Once $t$ is known, the horizontal distance follows from $s_x=u_{x}t$. This is the standard two-step structure: vertical for time, horizontal for range.

Launching at angle $\theta$ with speed $u$ means resolving first: $u_x=u\cos \theta$ and $u_y=u\sin \theta$. From there the method is unchanged.

Taking upwards as positive, the vertical acceleration is $-g$. At the highest point $v_y=0$, which gives the time to the top; for a projectile landing at its launch height, the flight is symmetric so the total time is twice that.

The range is then the horizontal velocity multiplied by the total time of flight.

Every calculation above neglects air resistance. In reality drag acts opposite to the velocity at all times and grows with speed, which changes the picture in several linked ways.

The horizontal velocity is no longer constant — it decreases throughout — so the range is shorter. The maximum height is lower. And the path loses its symmetry: the descent is steeper than the ascent, because the projectile is slower coming down and spends longer doing it.

Edexcel asks for this qualitatively rather than as a calculation, so the marks are for stating the effects and linking them to drag opposing motion.