M2.1 Projectiles

How do you model the motion of a projectile?

The modelling assumptions are: the body is a particle (size and spin ignored); there is no air resistance, so the only force after launch is the weight \(mg\); and \(g\) is constant at 9.8 m s⁻². So the acceleration is \(g\) vertically downwards, the same for every mass, and the path lies in the vertical plane containing the initial velocity.

A particle projected with speed \(u\) at angle \(\theta\) above the horizontal (the angle of projection) has initial velocity components \(u\cos\theta\) horizontally and \(u\sin\theta\) vertically.

The vertical column is just the M1 constant-acceleration formulae with \(a = -g\); \(v^2 = u^2 + 2as\) and \(s = \tfrac{1}{2}(u + v)t\) are available too. Time links the two directions: find \(t\) from one direction, then use it in the other.

How do you find the time of flight, range and greatest height?

At the greatest height the vertical velocity is zero: use \(v^2 = u^2 + 2as\) vertically with \(v = 0\). For the time of flight back to launch level, the vertical displacement is zero: use \(s = ut + \tfrac{1}{2}at^2\) with \(s = 0\) and take the non-zero root. The range is the horizontal distance travelled in that time. On level ground the path is symmetrical, so the time to the top is half the time of flight.

The general results (level ground)

From \(0 = (u\sin\theta)t - \tfrac{1}{2}gt^2\), the non-zero root is \(T = \dfrac{2u\sin\theta}{g}\). Then \(R = u\cos\theta \times \dfrac{2u\sin\theta}{g} = \dfrac{u^2\sin 2\theta}{g}\), using the double-angle identity from P3. From \(0 = (u\sin\theta)^2 - 2gH\), \(H = \dfrac{u^2\sin^2\theta}{2g}\). None of the three is in the formulae booklet, and they are valid only when the particle lands at the level it was launched from; a "show that" needs the derivation.

For a fixed speed, \(\sin 2\theta\) is greatest when \(2\theta = 90^\circ\), so the maximum range is \(u^2/g\), at \(\theta = 45^\circ\). Since \(\sin 2\theta = \sin(180^\circ - 2\theta)\), the angles \(\theta\) and \(90^\circ - \theta\) give the same range, with different heights and times.

How do you find the velocity at a given time?

The horizontal component is \(v_x = u\cos\theta\) (unchanged) and the vertical component is \(v_y = u\sin\theta - gt\): positive means moving upwards, negative means moving downwards. The speed is \(\sqrt{v_x^2 + v_y^2}\), and the direction makes angle \(\alpha\) with the horizontal where \(\tan\alpha = |v_y| \div v_x\), stated as above or below. The velocity is always along the tangent to the path. To find a speed at a given height without the time, use \(v^2 = u^2 + 2as\) vertically, then combine with the horizontal component.

What changes when a projectile is launched from a height?

Take the point of projection as the origin with upwards positive; landing \(h\) metres below the launch point means vertical displacement \(-h\). The time comes from a quadratic, \(-h = (u\sin\theta)t - 4.9t^2\); reject the negative root, which describes a time before the launch. The range formula does not apply, because launch and landing are at different levels. Projected horizontally, the initial vertical velocity is zero and \(-h = -4.9t^2\) gives the time directly; projected below the horizontal, the initial vertical component is \(-u\sin\theta\).

How do you aim a projectile at a target?

A target \(a\) m horizontally and \(b\) m vertically from the point of projection gives two conditions: \(x = a\) and \(y = b\) at the same time. With a known velocity, find \(t\) from the horizontal motion and substitute in the vertical equation. With an unknown speed or angle, eliminate \(t\) using \(t = a \div (u\cos\theta)\); this is where the trajectory equation comes from. The sign of \(v_y\) at that time tells you whether the particle is rising or falling, and to check whether it clears an obstacle, compare its height at the obstacle's horizontal distance with the obstacle's height.

What is the equation of the trajectory?

Eliminate \(t\) between the two displacements. From \(x = (u\cos\theta)t\), \(t = x \div (u\cos\theta)\); substitute into \(y = (u\sin\theta)t - \tfrac{1}{2}gt^2\):

\[y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta} = x\tan\theta - \frac{gx^2}{2u^2}\left(1 + \tan^2\theta\right).\]

The origin is the point of projection, with \(y\) vertically upwards; the second form uses \(1/\cos^2\theta = \sec^2\theta = 1 + \tan^2\theta\). It is not in the formulae booklet, and questions often ask you to show it. For fixed \(u\) and \(\theta\) this is \(y = px - qx^2\): the path is a parabola.

How do you find the angle of projection?

If the speed \(u\) is known and the path must pass through a point \((a, b)\), substitute into the second form of the trajectory equation. It becomes a quadratic in \(\tan\theta\):

\[\frac{ga^2}{2u^2}\tan^2\theta - a\tan\theta + \left(b + \frac{ga^2}{2u^2}\right) = 0.\]

Two positive roots give two possible angles: a low, fast-arriving path and a high, slow "lob". One repeated root means the point is just within reach at that speed; no real roots means it cannot be reached at speed \(u\). Derive this from the trajectory equation each time; learn the identity \(\sec^2\theta = 1 + \tan^2\theta\).

How do you describe a projectile with vectors?

Take \(\mathbf{i}\) and \(\mathbf{j}\) as unit vectors horizontally and vertically upwards. The initial velocity is \(\mathbf{u} = (u\cos\theta)\mathbf{i} + (u\sin\theta)\mathbf{j}\) and the acceleration is \(-g\mathbf{j}\). The vector constant-acceleration formulae from M1 give, with the origin at the point of projection,

\[\mathbf{r} = \mathbf{u}t - \tfrac{1}{2}gt^2\mathbf{j}, \qquad \mathbf{v} = \mathbf{u} - gt\mathbf{j}.\]

If the particle starts at position vector \(\mathbf{r}_0\), add \(\mathbf{r}_0\). The \(\mathbf{i}\)-component is the horizontal motion and the \(\mathbf{j}\)-component the vertical motion. "Moving parallel to \(\mathbf{i}\)" means the \(\mathbf{j}\)-component of \(\mathbf{v}\) is zero; "moving at 45° below the horizontal" means the \(\mathbf{j}\)-component equals minus the \(\mathbf{i}\)-component.

Exam tips for M2.1

Common misconceptions

How this topic connects

M1.2 Kinematics and M1.1 Vectors

The vertical motion is straight-line motion with constant acceleration, and vector form uses \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) and \(\mathbf{r} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2\).

M2.2 Variable acceleration

Differentiating \(\mathbf{r} = \mathbf{u}t - \tfrac{1}{2}gt^2\mathbf{j}\) gives \(\mathbf{v} = \mathbf{u} - gt\mathbf{j}\) and then \(\mathbf{a} = -g\mathbf{j}\): projectiles are the constant-acceleration case of calculus-based motion.

M2.4 Work, energy and power

With no air resistance, \(\tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2 = mgh\) gives the speed at any height, a fast check on worked example 4. P3 supplies \(\sec^2\theta = 1 + \tan^2\theta\) and \(\sin 2\theta = 2\sin\theta\cos\theta\).

Key formulae

Check your understanding

Specification checklist

• (M2 1.1) I can model a projectile as a particle moving in a vertical plane with a constant acceleration \(g\) downwards, and state the assumptions this makes.
• (M2 1.2) I can resolve the initial velocity into horizontal and vertical components and treat the two motions separately, linked by the time.
• (M2 1.2) I can find the time of flight, the range and the greatest height, and derive the general results for level ground.
• (M2 1.2) I can find the speed and direction of motion at a given time or position.
• (M2 1.2) I can solve problems where the projectile is launched from a height or must pass through or hit a given point.
• (M2 1.2) I can derive and use the equation of the trajectory, and form and solve a quadratic in \(\tan\theta\) to find possible angles of projection.
• (M2 1.2) I can write the position and velocity of a projectile as vectors in terms of \(\mathbf{i}\) and \(\mathbf{j}\) and use them to solve problems.