3.1 - Modelling projectile motion

The projectile model

If one ball is dropped and another is thrown horizontally from the same height at the same instant, which reaches level ground first? In the model we will use, they arrive together. Their horizontal motions differ, but their vertical motions start with the same vertical velocity and have the same acceleration.

Once a ball has left the hand, there is no continuing “throwing force”. Neglecting air resistance, its only force is its weight. Newton’s second law gives downward acceleration of magnitude $g$, independent of the ball’s mass. A downward acceleration does not mean downward velocity: the ball may still be moving upwards while its upward velocity decreases.

The model begins after release and ends at the first relevant impact. It does not describe the force exerted by the hand or the contact force during a bounce. Treating the body as a particle is useful when its dimensions are small compared with the distances involved. Neglecting air resistance is less suitable for a feather, a parachute or a rapidly moving light ball. Wind and spin can also make a real flight depart from this model.

Choose a horizontal axis and a vertical axis before calculating. With rightwards and upwards positive, the acceleration components are $a_x=0$ and $a_y=-g$. At the greatest height the vertical velocity is zero for an instant, but the acceleration is still $-g$. Unless another value is specified, use $g=10\,\mathrm{m\,s^{-2}}$ in the numerical work in these lessons.

[DIAGRAM: asset_name: Projectile velocity and acceleration; asset_slug: cam9231-projectile-components; caption: Horizontal velocity stays constant while vertical velocity changes sign; description: Exact trajectory x equals 20t and y equals 15t minus 5t squared from launch until t equals 3 seconds. Tangent arrows at t equals 1 and 2 show equal rightward components and opposite vertical components. A downward acceleration arrow at the highest point shows that gravity continues to act there.]

The two motions share the same clock. A horizontal launch has zero initial vertical velocity, just like a drop; adding a horizontal component cannot change the vertical arrival time.

Apply the model at its most easily misread instant.

Horizontal and vertical equations

Resolve the launch velocity first, then apply the constant-acceleration equations separately in the two directions.

A speed of $25\,\mathrm{m\,s^{-1}}$ at an angle $\theta$ above the horizontal, where $\cos\theta=4/5$ and $\sin\theta=3/5$, gives initial components $20$ rightwards and $15$ upwards, in $\mathrm{m\,s^{-1}}$. After $t$ seconds, the horizontal displacement is $20t$; the vertical displacement is $15t-5t^2$. The negative term represents the effect of downward acceleration, not a negative initial speed.

In general, take the launch point as the origin, with $x$ horizontal and $y$ vertically upwards. For initial speed $u$ at angle $\theta$ above the positive horizontal direction, resolving a right-angled triangle gives $u_x=u\cos\theta$ and $u_y=u\sin\theta$.

The same $t$ must appear in both equations: the particle cannot have a different elapsed time horizontally and vertically.

Change only the observation time. Doubling time doubles the horizontal displacement, but the vertical acceleration prevents the same rule vertically.