Projectiles

A projectile is a body moving freely under gravity alone. The whole topic rests on one idea: the horizontal and vertical motions are independent. Horizontally there is no force and the velocity is constant; vertically the acceleration is $-g$ throughout. Treat them as two separate one-dimensional problems joined only by the time $t$.

Take $g = 9.81\ \text{m s}^{-2}$ and upwards as positive.

Modelling assumptions

The standard model assumes

• no air resistance
• the body is a particle, so it does not spin and has no size
• $g$ is constant
• the ground is horizontal

Components of the initial velocity

For a launch speed $u$ at angle $\alpha$ above the horizontal:

$$u_x = u\cos\alpha \quad(\text{constant throughout}) \qquad u_y = u\sin\alpha$$

Then at time $t$:

$$x = (u\cos\alpha)t \qquad y = (u\sin\alpha)t - \tfrac12 gt^2$$

$$v_x = u\cos\alpha \qquad v_y = u\sin\alpha - gt$$

For $u = 30\ \text{m s}^{-1}$ at $40°$: $u_x = 22.98$ and $u_y = 19.28\ \text{m s}^{-1}$.

Time of flight and range

Over level ground the projectile returns to $y = 0$:

$$0 = (u\sin\alpha)t - \tfrac12 gt^2 = t\left(u\sin\alpha - \tfrac12 gt\right)$$

so $t = 0$ (the launch) or

$$T = \frac{2u\sin\alpha}{g}$$

The range is then the horizontal distance covered in that time:

$$R = u\cos\alpha \times T = \frac{u^2\sin 2\alpha}{g}$$

For the example above, $T = 3.93$ s and $R = 90.3$ m.

Maximum height

At the top $v_y = 0$, which happens at $t = \dfrac{u\sin\alpha}{g}$ — exactly half the time of flight. Then

$$H = \frac{u^2\sin^2\alpha}{2g}$$

For $u = 30$ at $40°$: $H = \dfrac{19.28^2}{19.62} = 18.95$ m.

The equation of the path

Eliminating $t$ between $x = (u\cos\alpha)t$ and the equation for $y$ gives

$$y = x\tan\alpha - \frac{gx^2}{2u^2\cos^2\alpha}$$

a downward parabola. Use it whenever a question links $x$ and $y$ without mentioning time — for instance "does the ball clear a 3 m wall 20 m away?"

Projection from a height

When the projectile starts above the ground, the flight does not end at $y = 0$ — it ends where $y$ equals the negative of the launch height.

A ball is thrown horizontally at $15\ \text{m s}^{-1}$ from a cliff 20 m high.

Horizontal projection means $u_y = 0$, so vertically it is a straight drop:

$$20 = \tfrac12(9.81)t^2 \quad\Longrightarrow\quad t = 2.02\ \text{s}$$

$$x = 15(2.02) = 30.3\ \text{m}$$

On landing, $v_y = 9.81(2.02) = 19.81\ \text{m s}^{-1}$ downwards, so

$$\text{speed} = \sqrt{15^2 + 19.81^2} = 24.8\ \text{m s}^{-1}, \qquad \theta = \tan^{-1}\frac{19.81}{15} = 52.9° \text{ below the horizontal}$$

Velocity at a given instant

Find each component, then recombine:

$$v = \sqrt{v_x^2 + v_y^2}, \qquad \theta = \tan^{-1}\frac{v_y}{v_x}$$

For $u = 30$ at $40°$ at $t = 2$ s:

$$v_x = 22.98 \qquad v_y = 19.28 - 9.81(2) = -0.34$$

$$v = 22.98\ \text{m s}^{-1} \text{ at } 0.8° \textbf{ below} \text{ the horizontal}$$

The negative $v_y$ says the projectile has just passed its highest point, which fits: the apex was at $t = 1.97$ s.