Short Notes: Motion in a Plane

3.1 Projectile Motion

Projectile motion is the motion of an object thrown or projected into the air, subject only to the acceleration of gravity. The motion can be analysed independently in horizontal and vertical directions by resolving initial velocity into components. The following table summarises the standard results for a projectile launched with initial speed \(u\) at an angle \(\theta\) above the horizontal, with gravitational acceleration \(g\) directed downwards.

Explanation and derivations

Resolve the initial velocity into horizontal and vertical components. Treat horizontal motion as uniform and vertical motion as uniformly accelerated.

Derivation of \(T\) (time of flight):

Vertical displacement for full flight is zero. Use \(y = u_y t - \tfrac{1}{2} g t^2\).

Set \(0 = u\sin\theta\; t - \tfrac{1}{2} g t^{2}\).

Non-zero solution for \(t\) gives \(\,T = \dfrac{2u\sin\theta}{g}\).

Derivation of \(H\) (maximum height):

At top, vertical velocity is zero. Use \(v_y^{2} = u_y^{2} - 2gH\).

Set \(0 = u^{2}\sin^{2}\theta - 2gH\).

Solve to get \(\,H = \dfrac{u^{2}\sin^{2}\theta}{2g}\).

Derivation of \(R\) (horizontal range):

Range is horizontal speed times total time: \(R = u_x T\).

Substitute \(u_x = u\cos\theta\) and \(T = \dfrac{2u\sin\theta}{g}\).

Simplify to obtain \(\,R = \dfrac{u^{2}\sin 2\theta}{g}\).

Trajectory shape and symmetry

The trajectory is a parabola given by the equation \(y = x\tan\theta - \dfrac{gx^{2}}{2u^{2}\cos^{2}\theta}\). The motion is symmetric about the highest point: the time to rise to the maximum height equals the time to descend from it.

Worked example

Example: A projectile is launched with speed \(u = 20\ \text{m s}^{-1}\) at angle \(\theta = 30^\circ\). Find the time of flight, maximum height and range. Take \(g = 9.8\ \text{m s}^{-2}\).

Compute the vertical component: \(u\sin\theta = 20\times\sin 30^\circ = 20\times 0.5 = 10\ \text{m s}^{-1}\).

Time of flight: \(T = \dfrac{2u\sin\theta}{g} = \dfrac{2\times 10}{9.8} = \dfrac{20}{9.8}\ \text{s} \approx 2.04\ \text{s}.\)

Maximum height: \(H = \dfrac{u^{2}\sin^{2}\theta}{2g} = \dfrac{(20)^{2}\times(0.5)^{2}}{2\times 9.8} = \dfrac{400\times 0.25}{19.6} = \dfrac{100}{19.6}\ \text{m} \approx 5.10\ \text{m}.\)

Range: horizontal speed \(u\cos\theta = 20\times\cos 30^\circ = 20\times 0.866 = 17.32\ \text{m s}^{-1}.\)

Range \(R = u_x T = 17.32\times 2.04 \approx 35.33\ \text{m}.\)

3.2 Circular Motion

Circular motion describes motion along a circular path. Quantities of interest include angular displacement, angular velocity, linear speed, centripetal acceleration, centripetal force, period and frequency. For motion in a circle of radius \(r\) with angular speed \(\omega\), the following relations hold.

Conceptual notes

Centripetal acceleration always points towards the centre of the circular path and changes only the direction of the velocity, not its magnitude. The force causing this acceleration is called the centripetal force; it may arise from tension, gravity, friction or normal reaction depending on the situation.

Derivation of centripetal acceleration (brief)

Consider uniform circular motion with constant speed \(v\) around a circle of radius \(r\). The change in velocity over a small time due to change in direction leads to acceleration directed towards the centre. Using geometry of infinitesimal angle \(d\theta\) and arc length \(ds = r\,d\theta\), one obtains \(a_c = \dfrac{v^{2}}{r} = r\omega^{2}\).

Worked example

Example: A stone tied to a string of length \(r = 0.5\ \text{m}\) is whirled in a horizontal circle at speed \(v = 10\ \text{m s}^{-1}\). Find the centripetal acceleration and the force on a stone of mass \(m = 0.2\ \text{kg}\).

Centripetal acceleration: \(a_c = \dfrac{v^{2}}{r} = \dfrac{100}{0.5} = 200\ \text{m s}^{-2}.\)

Centripetal force: \(F_c = m a_c = 0.2\times 200 = 40\ \text{N}.\)

3.3 Vector Operations

Vectors are quantities with both magnitude and direction. In plane motion, vectors represent displacement, velocity, acceleration and forces. Important vector operations and formulae are listed below.

• Magnitude: \(\,| \mathbf{A} | = \sqrt{A_x^{2} + A_y^{2} + A_z^{2}}\)
• Unit vector: \(\,\hat{\mathbf{A}} = \dfrac{\mathbf{A}}{|\mathbf{A}|}\)
• Dot product: \(\,\mathbf{A}\cdot\mathbf{B} = AB\cos\theta = A_xB_x + A_yB_y + A_zB_z\)
• Cross product (magnitude): \(\,|\mathbf{A}\times\mathbf{B}| = AB\sin\theta\) and direction given by the right-hand rule
• Resolution of a vector: \(A_x = A\cos\theta,\quad A_y = A\sin\theta\)
• Addition (resultant magnitude): \(\,R = \sqrt{A^{2} + B^{2} + 2AB\cos\theta}\)

Applications in motion in a plane

Vector resolution is essential to split velocities and accelerations into orthogonal components so that motion along perpendicular axes can be treated independently. The dot product is used to find the component of one vector along another and in work calculations. The cross product gives the area of the parallelogram spanned by two vectors and is used to find torque and angular momentum directions.

Example: resultant of two vectors

Example: Two vectors of magnitudes \(A = 5\) and \(B = 12\) act with an angle \(\theta = 60^\circ\) between them. Find the magnitude of the resultant.

Use the formula \(R = \sqrt{A^{2} + B^{2} + 2AB\cos\theta}\).

Compute \(R = \sqrt{5^{2} + 12^{2} + 2\times5\times12\times\cos 60^\circ}.\)

Since \(\cos 60^\circ = 0.5\), \(R = \sqrt{25 + 144 + 2\times5\times12\times 0.5}.\)

Evaluate \(R = \sqrt{25 + 144 + 60} = \sqrt{229} \approx 15.13.\)

End of short notes on motion in a plane. These formulae and methods form the core toolkit for analysing two-dimensional motion problems including projectiles, uniform circular motion and vector decomposition.