Projectile Motion

This topic explores fundamental concepts that shape our understanding of the world.

Projectile motion is the motion of a body launched into the air and subject only to the acceleration Due to gravity. By resolving the initial velocity into horizontal and vertical components and Applying the equations of motion independently in each direction, the complete trajectory can be Determined.

A projectile is launched with speed $V$ at angle $\theta$ to the horizontal from the origin.

Horizontal component: $V_x = V\cos\theta$

Vertical component: $V_y = V\sin\theta$

Taking upward as positive, with horizontal axis $x$ and vertical axis $y$:

Since there is no horizontal acceleration:

$\boxed{x = V\cos\theta \cdot t}$

$\dot{x} = V\cos\theta$

$\boxed{y = V\sin\theta \cdot t - \frac{1}{2}gt^2}$

$\dot{y} = V\sin\theta - gt$

$\ddot{y} = -g$

From the horizontal equation: $t = \dfrac{x}{V\cos\theta}$.

Substituting into the vertical equation:

$\boxed{y = x\tan\theta - \frac{gx^2}{2V^2\cos^2\theta}}$

Since this has the form $y = ax - bx^2$ (with $a = \tan\theta$ and $b = \frac{g}{2V^2\cos^2\theta}$), the trajectory is a parabola opening downward. $\blacksquare$

The projectile returns to $y = 0$ when:

$V\sin\theta \cdot t - \frac{1}{2}gt^2 = 0 \implies t(V\sin\theta - \frac{1}{2}gt) = 0$

$\boxed{T = \frac{2V\sin\theta}{g}}$

At maximum height, $\dot{y} = 0$:

$V\sin\theta - gt_{\mathrm{max}} = 0 \implies t_{\mathrm{max}} = \frac{V\sin\theta}{g}$

$\boxed{H = \frac{V^2\sin^2\theta}{2g}}$

This occurs at $x = V\cos\theta \cdot \dfrac{V\sin\theta}{g} = \dfrac{V^2\sin\theta\cos\theta}{g}$. $\blacksquare$

$R = V\cos\theta \cdot T = V\cos\theta \cdot \frac{2V\sin\theta}{g}$

$\boxed{R = \frac{V^2\sin 2\theta}{g}}$

This is maximised when $\sin 2\theta = 1$I.e., $\theta = 45^\circ$Giving $R_{\max} = \dfrac{V^2}{g}$. $\blacksquare$

A plane is inclined at angle $\alpha$ to the horizontal. A projectile is launched at angle $\theta$ Above the horizontal from the bottom of the plane.

The projectile lands on the plane when $y = x\tan\alpha$.

Setting $x\tan\alpha = x\tan\theta - \dfrac{gx^2}{2V^2\cos^2\theta}$:

$x\left(\tan\theta - \tan\alpha\right) = \frac{gx^2}{2V^2\cos^2\theta}$

$\boxed{x = \frac{2V^2\cos^2\theta(\tan\theta - \tan\alpha)}{g}}$

The range on the plane is $r = \dfrac{x}{\cos\alpha}$:

$\boxed{r = \frac{2V^2\cos\theta\sin(\theta - \alpha)}{g\cos^2\alpha}}$

When a projectile is launched from the top of a plane inclined at angle $\alpha$ below the Horizontal at angle $\theta$ above the horizontal, the landing condition is $y = -x\tan\alpha$:

$-x\tan\alpha = x\tan\theta - \frac{gx^2}{2V^2\cos^2\theta}$

$\boxed{r = \frac{2V^2\cos\theta\sin(\theta + \alpha)}{g\cos^2\alpha}}$

For maximum range up the plane, maximise $r = \dfrac{2V^2\cos\theta\sin(\theta-\alpha)}{g\cos^2\alpha}$.

Using the product-to-sum identity: $\cos\theta\sin(\theta-\alpha) = \frac{1}{2}[\sin(2\theta-\alpha) - \sin\alpha]$.

This is maximised when $\sin(2\theta - \alpha) = 1$Giving:

$2\theta - \alpha = 90° \implies \boxed{\theta = \frac{90° + \alpha}{2} = 45° + \frac{\alpha}{2}}$

For down the plane: $\theta = 45° - \dfrac{\alpha}{2}$. $\blacksquare$

An alternative approach is to take axes parallel and perpendicular to the plane. With $s$ along the Plane and $n$ perpendicular:

• Component of gravity along the plane: $g\sin\alpha$ (down the plane)
• Component of gravity perpendicular to the plane: $g\cos\alpha$ (into the plane)

The projectile lands on the plane when $n = 0$.

The velocity components at time $t$ are:

$v_x = V\cos\theta, \qquad v_y = V\sin\theta - gt$

The speed is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{V^2\cos^2\theta + (V\sin\theta - gt)^2}$.

The direction of motion is at angle $\phi$ to the horizontal where:

$\tan\phi = \frac{v_y}{v_x} = \frac{V\sin\theta - gt}{V\cos\theta} = \tan\theta - \frac{gt}{V\cos\theta}$

Starting from the trajectory equation, the projectile lands when $y = 0$:

$0 = R\tan\theta - \frac{gR^2}{2V^2\cos^2\theta}$

Either $R = 0$ (the launch point) or:

$\tan\theta = \frac{gR}{2V^2\cos^2\theta} = \frac{gR\sec^2\theta}{2V^2} = \frac{gR}{2V^2\cos^2\theta}$

Solving for $R$:

$R = \frac{2V^2\cos^2\theta\tan\theta}{g} = \frac{2V^2\sin\theta\cos\theta}{g} = \frac{V^2\sin 2\theta}{g}$

To maximise, differentiate with respect to $\theta$ and set to zero:

$\frac{dR}{d\theta} = \frac{V^2}{g}\cdot 2\cos 2\theta = 0 \implies \cos 2\theta = 0 \implies 2\theta = 90° \implies \theta = 45^\circ$

Second derivative check:

$\frac{d^2R}{d\theta^2} = \frac{V^2}{g}\cdot(-4\sin 2\theta) \bigg|_{\theta = 45°} = \frac{V^2}{g}(-4) \lt 0 \quad \checkmark$

So the maximum is confirmed. Substituting $\theta = 45^\circ$:

$R_{\max} = \frac{V^2\sin 90°}{g} = \frac{V^2}{g}$

A projectile is launched from height $h$ above ground level with speed $V$ at angle $\theta$ above The horizontal. Taking upward as positive with origin at the launch point:

$y = V\sin\theta \cdot t - \frac{1}{2}gt^2$

The projectile hits the ground when $y = -h$:

$V\sin\theta \cdot t - \frac{1}{2}gt^2 = -h$

$\frac{1}{2}gt^2 - V\sin\theta \cdot t - h = 0$

Using the quadratic formula (taking the positive root):

$\boxed{T = \frac{V\sin\theta + \sqrt{V^2\sin^2\theta + 2gh}}{g}}$

$R = V\cos\theta \cdot T = \frac{V\cos\theta\left(V\sin\theta + \sqrt{V^2\sin^2\theta + 2gh}\right)}{g}$

The maximum height above the launch point is unchanged from the ground-level case:

$H_{\mathrm{above launch}} = \frac{V^2\sin^2\theta}{2g}$

The maximum height above ground level is $h + \dfrac{V^2\sin^2\theta}{2g}$.

For maximum range from a height, the optimal angle is less than $45^\circ$. The exact value Satisfies:

$\theta = \arctan\!\left(\frac{V}{\sqrt{V^2 + 2gh}}\right)$

Maximise $R = V\cos\theta\cdot T$ where $T$ is given above. Equivalently, maximise:

$R(\theta) = \frac{V^2\sin\theta\cos\theta + V\cos\theta\sqrt{V^2\sin^2\theta + 2gh}}{g}$

Let $u = \sin\theta$. Then $\cos\theta = \sqrt{1 - u^2}$ and we maximise:

$R(u) \propto u\sqrt{1-u^2} + \sqrt{1-u^2}\sqrt{V^2 u^2 + 2gh}$

Differentiating and simplifying leads to the condition $\cos\theta = \dfrac{V}{\sqrt{V^2 + 2gh}}$I.e.:

$\tan\theta = \frac{V\sin\theta}{V\cos\theta} = \frac{V\sqrt{1 - \frac{V^2}{V^2 + 2gh}}}{\frac{V^2}{\sqrt{V^2 + 2gh}}} = \frac{V}{\sqrt{V^2 + 2gh}}$

When $h = 0$This reduces to $\tan\theta = 1$I.e., $\theta = 45^\circ$ as expected. $\blacksquare$

Example. A stone is thrown from a cliff $50\,\mathrm{m}$ high at $15\,\mathrm{m s}^{-1}$ at $30^\circ$ above the horizontal. Find the time of flight, the horizontal range, the maximum height Above ground, and the speed and direction of impact.

Time of flight:

$T = \frac{15\sin 30° + \sqrt{15^2\sin^2 30° + 2(9.8)(50)}}{9.8}$

$= \frac{7.5 + \sqrt{56.25 + 980}}{9.8} = \frac{7.5 + \sqrt{1036.25}}{9.8} = \frac{7.5 + 32.19}{9.8} \approx 4.05\,\mathrm{s}$

Range: $R = 15\cos 30° \times 4.05 \approx 12.99 \times 4.05 \approx 52.6\,\mathrm{m}$.

Maximum height above ground: $50 + \dfrac{15^2\sin^2 30°}{2(9.8)} = 50 + \dfrac{56.25}{19.6} \approx 50 + 2.87 = 52.87\,\mathrm{m}$.

Speed at impact:

$v_x = 15\cos 30° = 12.99\,\mathrm{m s}^{-1}$.

$v_y = 15\sin 30° - 9.8(4.05) = 7.5 - 39.69 = -32.19\,\mathrm{m s}^{-1}$.

Speed $= \sqrt{12.99^2 + 32.19^2} = \sqrt{168.7 + 1036.2} = \sqrt{1204.9} \approx 34.7\,\mathrm{m s}^{-1}$.

Angle below horizontal: $\arctan(32.19/12.99) \approx 68.1^\circ$.

Example. A projectile is launched at $30\,\mathrm{m s}^{-1}$ at $55^\circ$ to the horizontal up A plane inclined at $20^\circ$. Find the range on the plane and the time of flight.

Using the range formula:

$r = \frac{2V^2\cos\theta\sin(\theta - \alpha)}{g\cos^2\alpha} = \frac{2(900)\cos 55°\sin 35°}{9.8\cos^2 20°}$

$= \frac{1800 \times 0.5736 \times 0.5736}{9.8 \times 0.8830} = \frac{592.4}{8.653} \approx 68.5\,\mathrm{m}$

Time of flight: the projectile lands when $y = x\tan 20^\circ$.

From the trajectory equation:

$x = \frac{2V^2\cos^2\theta(\tan\theta - \tan\alpha)}{g} = \frac{2(900)\cos^2 55°(\tan 55° - \tan 20°)}{9.8}$

$= \frac{1800 \times 0.3290 \times (1.4281 - 0.3640)}{9.8} = \frac{1800 \times 0.3290 \times 1.0641}{9.8} \approx 64.3\,\mathrm{m}$

$T = \frac{x}{V\cos\theta} = \frac{64.3}{30\cos 55°} = \frac{64.3}{17.21} \approx 3.74\,\mathrm{s}$

The horizontal distance at landing is $x = \dfrac{2V^2\cos^2\theta(\tan\theta - \tan\alpha)}{g}$.

Since $x = V\cos\theta \cdot T$:

$\boxed{T = \frac{2V\cos\theta(\tan\theta - \tan\alpha)}{g} = \frac{2V\sin(\theta - \alpha)}{g\cos\alpha}}$

Similarly:

$\boxed{T = \frac{2V\cos\theta(\tan\theta + \tan\alpha)}{g} = \frac{2V\sin(\theta + \alpha)}{g\cos\alpha}}$

The most common error in projectile motion is inconsistent sign conventions. If you define upward as Positive, then:

• $g$ appears as $-g$ in the acceleration, giving $y = V\sin\theta\cdot t - \frac{1}{2}gt^2$
• A projectile landing below the launch point has $y = -h$ at impact, not $y = h$
• The final vertical velocity is negative when the projectile is moving downward

If you define downward as positive, then $g$ is positive but $V\sin\theta$ becomes negative for Upward projection. Pick one convention and stick with it throughout the entire problem.

When working with inclined planes, the angle $\alpha$ is the angle of the plane to the Horizontal, not the angle of projection. Common mistakes:

• Confusing $\theta$ (projection angle) with $\alpha$ (plane angle)
• Using $\theta - \alpha$ for the down-the-plane case (should be $\theta + \alpha$)
• Forgetting that the range formula $r = x/\cos\alpha$ converts horizontal distance to distance along the plane

Two angles $\theta$ and $90° - \theta$ give the same range but different trajectories. The Steeper angle:

• Reaches a greater maximum height
• Has a longer time of flight
• Has a smaller horizontal component of velocity at every point

If an exam question asks about the trajectory (height, time, speed at a specific point), the Complementary angle will give a different answer even though the range is the same.

Always check that your answer makes physical sense:

• Range should be positive
• Time of flight should be positive
• The speed at impact from a height must exceed the launch speed (energy gained from gravity)
• The angle of impact should be steeper than the angle of projection (for horizontal ground launches)

Problem. A particle is projected up a plane inclined at $30°$ to the horizontal with speed $20\,\mathrm{m\,s^{-1}}$ at an angle of $50°$ to the horizontal. Find the range along the plane.

Solution. Resolving perpendicular to the plane (call this the $\xi$-axis) and parallel to the Plane (the $\eta$-axis):

a_\xi = -g\cos 30° = -\dfrac{g\sqrt{3}}{2}$$a_\eta = -g\sin 30° = -\dfrac{g}{2}.

$u_\xi = 20\sin(50° - 30°) = 20\sin 20° \approx 6.84\,\mathrm{m\,s^{-1}}$.

$u_\eta = 20\cos 20° \approx 18.79\,\mathrm{m\,s^{-1}}$.

The particle lands when $\xi = 0$ again:

$\xi = u_\xi t + \dfrac{1}{2}a_\xi t^2 = 0 \implies t\!\left(20\sin 20° - \dfrac{g\sqrt{3}}{4}\,t\right) = 0$.

Time of flight: $T = \dfrac{80\sin 20°}{g\sqrt{3}} \approx \dfrac{27.36}{17.06} \approx 1.604\,\mathrm{s}$.

Range along plane: $\eta = u_\eta T + \dfrac{1}{2}a_\eta T^2 = 20\cos 20° \times 1.604 - \dfrac{9.8}{2}(1.604)^2$

$\approx 30.14 - 12.60 = \boxed{17.5\,\mathrm{m}}$ (along the incline).

Problem. Show that the angle of projection $\theta$ for maximum range $R$ up a plane of Inclination $\alpha$ satisfies $\theta = \dfrac{\pi}{4} + \dfrac{\alpha}{2}$.

Solution. The range formula for a plane inclined at angle $\alpha$ is:

$R = \frac{2u^2\cos\theta\sin(\theta - \alpha)}{g\cos^2\alpha}$

Using $\sin A\cos B = \dfrac{1}{2}[\sin(A+B) + \sin(A-B)]$:

$R = \frac{u^2[\sin(2\theta - \alpha) - \sin\alpha]}{g\cos^2\alpha}$

$R$ is maximised when $\sin(2\theta - \alpha) = 1$I.e., $2\theta - \alpha = \dfrac{\pi}{2}$.

$\boxed{\theta = \frac{\pi}{4} + \frac{\alpha}{2}}$

Problem. A particle is projected from the origin with speed $u$ at angle $\theta$ above the Horizontal. At the same instant, a second particle is released from rest at position $(d, h)$. Find The condition on $u$ and $\theta$ for a collision.

Solution. The second particle falls freely: x_2(t) = d$$y_2(t) = h - \dfrac{1}{2}gt^2.

The first particle: x_1(t) = u\cos\theta\,t$$y_1(t) = u\sin\theta\,t - \dfrac{1}{2}gt^2.

For collision: $u\cos\theta\,t = d \implies t = \dfrac{d}{u\cos\theta}$.

Then: $u\sin\theta \cdot \dfrac{d}{u\cos\theta} - \dfrac{1}{2}g\!\left(\dfrac{d}{u\cos\theta}\right)^{\!2} = h$.

$d\tan\theta - \frac{gd^2}{2u^2\cos^2\theta} = h$

$\boxed{u^2 = \frac{gd^2}{2\cos^2\theta\,(d\tan\theta - h)}}$

Provided $d\tan\theta > h$.

Problem. A particle of mass $m$ is projected vertically upward at speed $u$. The air resistance Is $mkv^2$ opposing motion. Find the maximum height.

Solution. Going up: $\dfrac{dv}{dt} = -g - kv^2$.

$\int_0^u \frac{v\,dv}{g + kv^2} = \int_0^H dh$

Let w = g + kv^2$$dw = 2kv\,dv:

$\frac{1}{2k}\int_g^{g+ku^2} \frac{dw}{w} = \frac{1}{2k}\ln\!\left(\frac{g+ku^2}{g}\right) = H$

$\boxed{H = \frac{1}{2k}\ln\!\left(1 + \frac{ku^2}{g}\right)}$

Problem. A projectile has position $(x, y)$ at time $t$ given by $x = V\cos\theta\,t$ and $y = V\sin\theta\,t - \dfrac{1}{2}gt^2$. Derive the Cartesian equation and identify the key Features.

Solution. Eliminating $t$: $t = \dfrac{x}{V\cos\theta}$.

$y = x\tan\theta - \frac{gx^2}{2V^2\cos^2\theta} = x\tan\theta - \frac{gx^2\sec^2\theta}{2V^2}$

$\boxed{y = x\tan\theta - \frac{gx^2}{2V^2}(1 + \tan^2\theta)}$

This is a parabola. Setting $y = 0$: $x = 0$ or $x = \dfrac{2V^2\sin\theta\cos\theta}{g} = \dfrac{V^2\sin 2\theta}{g}$ (the Range).

Maximum height: $y_{\max} = \dfrac{V^2\sin^2\theta}{2g}$ at $x = \dfrac{V^2\sin 2\theta}{2g}$.

Problem. A gun can fire a shell with speed $u$ at any angle. Show that no point outside the Parabola $y = \dfrac{u^2}{2g} - \dfrac{gx^2}{2u^2}$ can be hit.

Solution. For angle $\theta$The trajectory is $y = x\tan\theta - \dfrac{gx^2}{2u^2}(1+\tan^2\theta)$.

Rearranging as a quadratic in $\tan\theta$:

$\frac{gx^2}{2u^2}\tan^2\theta - x\tan\theta + \frac{gx^2}{2u^2} + y = 0$

For a real angle to exist, the discriminant must be $\geq 0$:

$x^2 - 4 \cdot \frac{gx^2}{2u^2}\!\left(\frac{gx^2}{2u^2} + y\right) \geq 0$

$x^2 - \frac{2gx^2}{u^2}\!\left(\frac{gx^2}{2u^2} + y\right) \geq 0$

$1 - \frac{2g}{u^2}\!\left(\frac{gx^2}{2u^2} + y\right) \geq 0 \implies y \leq \frac{u^2}{2g} - \frac{gx^2}{2u^2}$

$\blacksquare$

A cricketer hits a ball from ground level with speed $25\,\mathrm{m\,s^{-1}}$ at $35°$ to the Horizontal. The ball just clears a wall $5\,\mathrm{m}$ high. Find the distance from the batsman to The wall.

Prove that the time of flight of a projectile on a plane inclined at angle $\alpha$ below the Horizontal is $T = \dfrac{2u\sin(\theta+\alpha)}{g\cos\alpha}$.

A particle is projected from a point $A$ on a cliff $40\,\mathrm{m}$ above sea level. It lands in The sea at a horizontal distance of $100\,\mathrm{m}$ from the foot of the cliff. If the angle of Projection is $30°$ above the horizontal, find the initial speed.

Both topics involve resolving forces and using Newton”s second law in 2D. See Circular Motion.

The trajectory equation is derived by eliminating the parameter $t$ from the parametric equations, a Standard calculus technique. See Further Calculus.

Conservation of energy provides an alternative to resolving forces, connecting projectiles to the Work-energy principle.

A ball is thrown from a height of $1.5\,\mathrm{m}$ at $10\,\mathrm{m\,s^{-1}}$ at $30°$ above the Horizontal. Find: (a) the time to reach maximum height; (b) the maximum height above the ground; (c) The horizontal range (distance from launch to landing).