Derivation

Projectile Motion from Constant-Acceleration Kinematics

D-420 Home PU-101 Threads force · energy Depends on Galilean Invariance of Newton's Laws, The Work-Energy Theorem

Statement

For a particle of constant mass \(m\) moving under a constant acceleration \(\vec{a}\), the position and velocity are linear and quadratic functions of time: \(\vec{v}(t)=\vec{v}_0+\vec{a}t\) and \(\vec{r}(t)=\vec{r}_0+\vec{v}_0 t+\tfrac{1}{2}\vec{a}t^2\). When this acceleration is the uniform gravitational field \(\vec{a}=-g\,\hat{y}\) near the Earth's surface and air resistance is neglected, a projectile launched with speed \(v_0\) at angle \(\theta\) above the horizontal traces the parabola \(y(x)=x\tan\theta-\dfrac{g\,x^2}{2v_0^2\cos^2\theta}\), with time of flight \(T=\dfrac{2v_0\sin\theta}{g}\), maximum height \(H=\dfrac{v_0^2\sin^2\theta}{2g}\), and range on level ground \(R=\dfrac{v_0^2\sin 2\theta}{g}\).

Why it matters

Projectile motion is the first two-dimensional kinematics problem in which the two axes decouple completely. Because the horizontal and vertical accelerations are independent, a single constant-acceleration calculation on each axis fully determines the trajectory. This decoupling is the conceptual template for every later separation of variables in physics: from simple harmonic motion to the Schrödinger equation.

It also gives the cleanest possible introduction to the idea that a vector equation of motion can be split into independent scalar equations. The parabolic trajectory, the \(45^\circ\) range optimum, and the complementary-angle symmetry are all consequences of just two integrations of a constant, making this the ideal bridge from algebraic kinematics to the calculus-based dynamics that follows.

Assumptions

The acceleration \(\vec{a}\) is constant in both magnitude and direction.If \(\vec{a}\) depends on position, velocity, or time (air drag, a spring, a varying gravitational field), the integrations below no longer yield the simple linear/quadratic forms; one must solve a differential equation instead.

The gravitational field is uniform: \(\vec{g}=-g\,\hat{y}\) with \(g=9.8\ \mathrm{m/s^2}\) constant.Over large vertical distances \(g\) decreases as \(1/r^2\), so the constant-\(g\) kinematics fail and the trajectory is no longer a parabola but a segment of an ellipse.

The launch and landing points are at the same height (level ground) for the range formula.If the projectile lands at a different elevation, the time of flight and range change; the neat \(R=v_0^2\sin 2\theta/g\) no longer holds.

The reference frame is inertial and the Earth's rotation is negligible.On a rotating Earth, a Coriolis acceleration \(-2\vec{\Omega}\times\vec{v}\) acts on the projectile, deflecting it eastward or westward and breaking the pure parabolic symmetry; for long-range artillery this correction is measurable.

The projectile is a point mass with no aerodynamic forces.A real object experiences drag \(\propto v^2\) and lift, which couple the horizontal and vertical motions, shorten the range, and make the descent steeper than the ascent; the symmetric parabola is then only an approximation.

Derivation

\[ \vec{a} = \frac{d\vec{v}}{dt} = \text{constant} \]

Definition of acceleration as the time derivative of velocity; the hypothesis is that this derivative is a fixed vector. A

\[ \int_{\vec{v}_0}^{\vec{v}(t)} d\vec{v} = \int_{0}^{t} \vec{a}\,dt \quad\Longrightarrow\quad \vec{v}(t) = \vec{v}_0 + \vec{a}t \]

Integrate the constant acceleration over time; the left side is the change in velocity. A

\[ \vec{v} = \frac{d\vec{r}}{dt} \]

Definition of velocity as the time derivative of position; this links the velocity found in Step 2 to the trajectory. A

\[ \int_{\vec{r}_0}^{\vec{r}(t)} d\vec{r} = \int_{0}^{t}\left(\vec{v}_0+\vec{a}t'\right)dt' \]

Substitute the velocity from Step 2 into the velocity definition; the dummy variable \(t'\) avoids confusion with the upper limit. A

\[ \vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2}\vec{a}t^2 \]

Evaluate the integral: \(\int_0^t \vec{v}_0\,dt' = \vec{v}_0 t\) and \(\int_0^t \vec{a}t'\,dt' = \tfrac{1}{2}\vec{a}t^2\). The factor \(\tfrac12\) is the hallmark of integrating a linear function. B

\[ v^2 = v_0^2 + 2a(x-x_0) \]

Eliminate time between Steps 2 and 5 (or integrate \(a\,dx=v\,dv\)); this is the time-independent kinematic relation. B

\[ a_x = 0,\qquad a_y = -g \]

For a projectile in uniform gravity with no air resistance, the only acceleration is vertical; the horizontal component vanishes. A

\[ x(t) = x_0 + v_0\cos\theta\; t,\qquad y(t) = y_0 + v_0\sin\theta\; t - \frac{1}{2}g t^2 \]

Apply Steps 2 and 5 to each axis with \(v_{0x}=v_0\cos\theta\), \(v_{0y}=v_0\sin\theta\), and \(a_x=0\), \(a_y=-g\). A

\[ t = \frac{x}{v_0\cos\theta} \quad\Longrightarrow\quad y(x) = x\tan\theta - \frac{g\,x^2}{2v_0^2\cos^2\theta} \]

Solve the horizontal equation for \(t\) (valid since \(a_x=0\)) and substitute into the vertical equation; the result is a quadratic in \(x\), i.e. a parabola. C

\[ y(T)=0 \;\Longrightarrow\; T = \frac{2v_0\sin\theta}{g} \]

Set \(y(T)=0\) for level ground (\(y_0=0\)); the non-trivial root of the quadratic in \(t\) gives the time of flight. A

\[ H = \frac{v_0^2\sin^2\theta}{2g} \]

The apex occurs when \(v_y=0\), i.e. \(t^*=v_0\sin\theta/g\); substituting into \(y(t)\) gives the maximum height. B

\[ R = x(T) = \frac{v_0^2\sin 2\theta}{g} \]

Evaluate the horizontal position at \(t=T\); using \(\sin 2\theta = 2\sin\theta\cos\theta\) gives the range on level ground. B

\[ \frac{dR}{d\theta} = \frac{2v_0^2\cos 2\theta}{g} = 0 \;\Longrightarrow\; \theta = 45^\circ \]

Maximise \(R\) with respect to \(\theta\); \(\cos 2\theta=0\) gives \(2\theta=90^\circ\), so \(\theta=45^\circ\) is the optimal launch angle. B

\[ R(\theta) = R(90^\circ-\theta) \]

Since \(\sin 2\theta = \sin(180^\circ-2\theta) = \sin 2(90^\circ-\theta)\), complementary launch angles give identical range. A

Result

\[ y(x) = x\tan\theta - \frac{g\,x^2}{2v_0^2\cos^2\theta},\qquad T = \frac{2v_0\sin\theta}{g},\qquad H = \frac{v_0^2\sin^2\theta}{2g},\qquad R = \frac{v_0^2\sin 2\theta}{g} \]

Reading. The trajectory is a downward-opening parabola in \(x\). The time of flight depends only on the vertical component of the initial velocity. The maximum height depends only on the vertical component. The range depends on both components through \(\sin 2\theta\), which is maximised at \(\theta=45^\circ\). Complementary angles \(\theta\) and \(90^\circ-\theta\) give the same range but different flight times and heights.

Units check. In \(y(x)\): \([x\tan\theta]=\mathrm{m}\), \([gx^2/v_0^2]=\mathrm{(m/s^2)(m^2)/(m^2/s^2)=m}\). In \(T\): \([v_0\sin\theta/g]=\mathrm{(m/s)/(m/s^2)=s}\). In \(H\): \([v_0^2/g]=\mathrm{(m^2/s^2)/(m/s^2)=m}\). In \(R\): \([v_0^2/g]=\mathrm{m}\). All consistent.

Limiting cases

Vertical throw (\(\theta=90^\circ\)). \(R=0\), \(H=v_0^2/(2g)\), \(T=2v_0/g\). The projectile goes straight up and down; the horizontal equation is trivially \(x=0\).
Horizontal launch (\(\theta=0\)) from height \(h\). \(T=\sqrt{2h/g}\), \(R=v_0\sqrt{2h/g}\). The vertical motion is pure free fall; the horizontal motion is uniform.
Complementary angles. \(\theta\) and \(90^\circ-\theta\) give the same \(R\) but different \(T\) and \(H\): the steeper angle has longer flight time and greater height.
Small-angle limit (\(\theta\ll 1\)). \(R\approx 2v_0^2\theta/g\), \(H\approx v_0^2\theta^2/(2g)\); the trajectory is nearly a straight line with slight downward curvature.

Breaks when

Air resistance. Drag \(\propto v^2\) opposes the velocity, coupling the horizontal and vertical equations. The trajectory is no longer a parabola; the descent is steeper than the ascent, and the range is shorter than the vacuum prediction.
Large vertical distances. When the altitude is a significant fraction of the Earth's radius, \(g\) decreases as \(1/r^2\) and the trajectory is a segment of a Keplerian ellipse, not a parabola. The constant-\(g\) approximation fails.
Earth's rotation (Coriolis effect). In the rotating frame of the Earth, a Coriolis acceleration \(-2\vec{\Omega}\times\vec{v}\) deflects the projectile. For long-range trajectories (artillery, intercontinental missiles) this produces measurable deflections, to the right of the direction of motion in the Northern Hemisphere and to the left in the Southern.
Non-level ground. If the landing point is at a different elevation than the launch point, the time of flight and range formulas change; the symmetric parabola about the apex no longer intersects the ground at equal \(y\).

Failure modes

Confusing the range formula with the height formula. Writing \(R=v_0^2\sin^2\theta/g\) (missing the factor of 2 and using \(\sin^2\) instead of \(\sin 2\theta\)). The range involves \(\sin 2\theta=2\sin\theta\cos\theta\), not \(\sin^2\theta\).
Forgetting that horizontal velocity is constant. Trying to apply the vertical kinematic equations to the horizontal motion, or vice versa. Since \(a_x=0\), \(v_x\) is constant and \(x(t)\) is linear in \(t\); the \(\tfrac12 at^2\) term appears only in \(y(t)\).
Using the wrong angle in the components. Assigning \(v_{0x}=v_0\sin\theta\) and \(v_{0y}=v_0\cos\theta\). The angle \(\theta\) is measured from the horizontal, so the horizontal component is \(v_0\cos\theta\) and the vertical component is \(v_0\sin\theta\).
Assuming the time to reach the apex equals the total flight time. The time to the apex is \(t^*=v_0\sin\theta/g\), which is exactly half the total flight time \(T=2v_0\sin\theta/g\) only for level ground. On uneven ground, the two are not related by a factor of 2.

Discussion

The power of the constant-acceleration kinematics lies in their universality. The same three equations — \(v=v_0+at\), \(x=x_0+v_0t+\tfrac12at^2\), \(v^2=v_0^2+2a\Delta x\) — describe a car braking on a straight road, a ball falling from a tower, and a projectile arcing through the air. The only difference is the value and direction of \(\vec{a}\). This universality is a direct consequence of the acceleration being constant: the integrals are elementary, and no differential equation needs to be solved.

The decoupling of horizontal and vertical motion is the key conceptual insight. Because \(a_x=0\) and \(a_y=-g\) are independent, the two axes can be treated as separate one-dimensional problems. The horizontal motion is uniform (constant velocity), and the vertical motion is uniformly accelerated (constant acceleration). The trajectory emerges from the superposition of these two independent motions, and the parabolic shape is a geometric consequence of one axis being linear in time and the other quadratic in time.

The \(45^\circ\) optimum and the complementary-angle symmetry are elegant consequences of the \(\sin 2\theta\) dependence of the range. The function \(\sin 2\theta\) is maximised at \(2\theta=90^\circ\), giving \(\theta=45^\circ\). The symmetry \(\sin 2\theta=\sin 2(90^\circ-\theta)\) means that launching at \(30^\circ\) and at \(60^\circ\) with the same speed gives the same range, but the \(60^\circ\) trajectory is higher and longer in flight time. This symmetry is a direct reflection of the interchangeability of the sine and cosine in the range formula.

At a deeper level, the parabola is an approximation to an ellipse. Under the true \(1/r^2\) force a thrown stone is a bound body (its energy is far below escape), so its path is a very elongated Keplerian ellipse with the Earth’s centre at the far focus; the small arc above the ground is indistinguishable from a parabola because over heights and distances small compared with \(R_\oplus = 6371\ \mathrm{km}\) the field is uniform to within a fraction \(2h/R_\oplus\) and parallel to within an angle \(x/R_\oplus\). The same approximation underlies the flat-Earth, uniform-field picture used throughout introductory mechanics.

Common misconceptions. The most persistent is the belief that a projectile in flight is “pushed” by some force in the direction of motion. In fact, after launch, the only force is gravity (neglecting air resistance), which acts vertically downward. The horizontal motion continues at constant velocity because no horizontal force acts. Another misconception is that the projectile is “at rest” at the top of its trajectory; in fact, it has zero vertical velocity but non-zero horizontal velocity, so it is still moving. The speed is minimum at the apex, not zero.

Worked examples

Example 1. A ball is launched with speed \(v_0=20\ \mathrm{m/s}\) at an angle \(\theta=30^\circ\) above the horizontal from ground level. Find the time of flight, maximum height, and range. Take \(g=9.8\ \mathrm{m/s^2}\).

\[ T = \frac{2v_0\sin\theta}{g} = \frac{2(20)\sin 30^\circ}{9.8} = \frac{40(0.5)}{9.8} = \frac{20}{9.8} \approx 2.04\ \mathrm{s} \]

Compute the time of flight using the formula; \(\sin 30^\circ=0.5\). A

\[ H = \frac{v_0^2\sin^2\theta}{2g} = \frac{(20)^2(0.5)^2}{2(9.8)} = \frac{400(0.25)}{19.6} = \frac{100}{19.6} \approx 5.10\ \mathrm{m} \]

Compute the maximum height; \(\sin^2 30^\circ=0.25\). A

\[ R = \frac{v_0^2\sin 2\theta}{g} = \frac{(20)^2\sin 60^\circ}{9.8} = \frac{400(0.866)}{9.8} = \frac{346.4}{9.8} \approx 35.3\ \mathrm{m} \]

Compute the range; \(\sin 60^\circ=0.866\). A

\[ T \approx 2.04\ \mathrm{s},\qquad H \approx 5.10\ \mathrm{m},\qquad R \approx 35.3\ \mathrm{m} \]

Reading. The ball is in the air for about 2 seconds, reaches a height of about 5 metres, and travels about 35 metres horizontally. The range is less than the \(45^\circ\) optimum because \(30^\circ\) is not the optimal angle.

Units check. \([T]=\mathrm{s}\), \([H]=\mathrm{m}\), \([R]=\mathrm{m}\). All correct.

Example 2. A projectile is launched from a cliff of height \(h=50\ \mathrm{m}\) with speed \(v_0=15\ \mathrm{m/s}\) at an angle \(\theta=45^\circ\) above the horizontal. Find the time of flight and the horizontal range. Take \(g=9.8\ \mathrm{m/s^2}\).

\[ y(t) = 50 + 15\sin 45^\circ\; t - \frac{1}{2}(9.8)t^2 = 50 + 10.61\,t - 4.9\,t^2 \]

Write the vertical position equation with \(y_0=50\ \mathrm{m}\), \(v_{0y}=15\sin 45^\circ=10.61\ \mathrm{m/s}\). A

\[ 0 = 50 + 10.61\,T - 4.9\,T^2 \quad\Longrightarrow\quad 4.9\,T^2 - 10.61\,T - 50 = 0 \]

Set \(y(T)=0\) for landing at ground level; rearrange into standard quadratic form. B

\[ T = \frac{10.61 + \sqrt{(10.61)^2 + 4(4.9)(50)}}{2(4.9)} = \frac{10.61 + \sqrt{112.6 + 980}}{9.8} = \frac{10.61 + \sqrt{1092.6}}{9.8} \] \[ T = \frac{10.61 + 33.05}{9.8} = \frac{43.66}{9.8} \approx 4.46\ \mathrm{s} \]

Solve the quadratic; take the positive root. B

\[ R = x(T) = v_0\cos\theta\; T = 15\cos 45^\circ\;(4.45) = 10.61(4.46) \approx 47.3\ \mathrm{m} \]

Compute the horizontal range using the constant horizontal velocity. A

\[ T \approx 4.46\ \mathrm{s},\qquad R \approx 47.3\ \mathrm{m} \]

Reading. The projectile is in the air for about 4.5 seconds and travels about 47 metres horizontally. The flight time is longer than the level-ground case because the projectile starts from a height, giving it more time to fall.

Units check. \([T]=\mathrm{s}\), \([R]=\mathrm{m}\). All correct.

Problems

A ball is thrown straight upward with speed \(v_0=15\ \mathrm{m/s}\). Find the maximum height and the total time of flight. Take \(g=9.8\ \mathrm{m/s^2}\).
Solution
For \(\theta=90^\circ\): \(H=v_0^2/(2g)=(15)^2/(2(9.8))=225/19.6\approx 11.5\ \mathrm{m}\). \(T=2v_0/g=2(15)/9.8=30/9.8\approx 3.06\ \mathrm{s}\).
A projectile is launched with speed \(v_0=30\ \mathrm{m/s}\) at an angle \(\theta=60^\circ\) above the horizontal from ground level. Find the time of flight, maximum height, and range. Take \(g=9.8\ \mathrm{m/s^2}\).
Solution
\(T=2(30)\sin 60^\circ/9.8=60(0.866)/9.8=51.96/9.8\approx 5.30\ \mathrm{s}\). \(H=(30)^2\sin^2 60^\circ/(2(9.8))=900(0.75)/19.6=675/19.6\approx 34.4\ \mathrm{m}\). \(R=(30)^2\sin 120^\circ/9.8=900(0.866)/9.8=779.4/9.8\approx 79.5\ \mathrm{m}\).
Show that the range \(R=v_0^2\sin 2\theta/g\) is maximised at \(\theta=45^\circ\). What is the maximum range for \(v_0=25\ \mathrm{m/s}\)?
Solution
\(dR/d\theta=2v_0^2\cos 2\theta/g=0\Rightarrow\cos 2\theta=0\Rightarrow 2\theta=90^\circ\Rightarrow\theta=45^\circ\). \(R_{\max}=v_0^2\sin 90^\circ/g=(25)^2(1)/9.8=625/9.8\approx 63.8\ \mathrm{m}\).
A projectile is launched from a cliff of height \(h=20\ \mathrm{m}\) with speed \(v_0=20\ \mathrm{m/s}\) at an angle \(\theta=30^\circ\) above the horizontal. Find the time of flight and the horizontal range. Take \(g=9.8\ \mathrm{m/s^2}\).
Solution
\(y(t)=20+20\sin 30^\circ\;t-4.9t^2=20+10t-4.9t^2\). Set \(y(T)=0\): \(4.9T^2-10T-20=0\). \(T=[10+\sqrt{100+392}]/9.8=[10+\sqrt{492}]/9.8=[10+22.18]/9.8=32.18/9.8\approx 3.28\ \mathrm{s}\). \(R=20\cos 30^\circ\;(3.284)=17.32(3.284)\approx 56.9\ \mathrm{m}\).
Two projectiles are launched with the same speed \(v_0\) at complementary angles \(\theta\) and \(90^\circ-\theta\). Show that they have the same range but different maximum heights and flight times. For \(v_0=20\ \mathrm{m/s}\), \(\theta=30^\circ\), compute both sets of values.
Solution
\(R(\theta)=v_0^2\sin 2\theta/g\). \(R(90^\circ-\theta)=v_0^2\sin(180^\circ-2\theta)/g=v_0^2\sin 2\theta/g=R(\theta)\). Heights: \(H(30^\circ)=(20)^2(0.5)^2/(2(9.8))=100/19.6\approx 5.10\ \mathrm{m}\); \(H(60^\circ)=(20)^2(0.866)^2/(2(9.8))=400(0.75)/19.6=300/19.6\approx 15.3\ \mathrm{m}\). Times: \(T(30^\circ)=2(20)(0.5)/9.8=20/9.8\approx 2.04\ \mathrm{s}\); \(T(60^\circ)=2(20)(0.866)/9.8=34.64/9.8\approx 3.53\ \mathrm{s}\). Same range, different heights and times.