Projectile Motion

Overview

Projectile motion is two-dimensional motion under gravity where an object is launched into the air and then moves freely after release.

This topic builds on:

• Kinematics
• Vectors
• Constant Acceleration Models

The key method is to split motion into:

• horizontal motion
• vertical motion

and solve them separately using a common time variable.

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Core Ideas

What is Projectile Motion?

A projectile is an object that, after launch, moves under the influence of gravity alone (ideal model).

Examples:

• thrown ball
• kicked football
• launched stone
• package dropped from aircraft

Its path is usually parabolic when air resistance is neglected.

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Assumptions

Unless stated otherwise, use the standard H2 model:

• air resistance is negligible
• acceleration due to gravity is constant
• gravity acts vertically downward
• Earth curvature is ignored
• projectile treated as a particle

Hence:

$$a_x=0$$ $$a_y=-g$$

if upward is chosen as positive.

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Resolving the Initial Velocity

If launch speed is $u$ at angle $\theta$ above horizontal:

$$u_x=u\cos \theta$$ $$u_y=u\sin \theta$$

Figure: Resolve the launch velocity into horizontal and vertical components. The horizontal component stays constant, while the vertical component changes under gravity.

These become the initial velocities in each direction.

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Horizontal and Vertical Motion

Horizontal Motion

Since no horizontal acceleration:

$$a_x=0$$

So:

$$v_x=u_x$$ $$s_x=u_{x}t$$

Horizontal velocity remains constant.

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Vertical Motion

Take upward as positive:

$$a_y=-g$$

where $g=9.81{\text{m s}}^{-2}$ is the magnitude of gravitational acceleration.

Using constant-acceleration equations:

$$v_y=u_y-gt$$ $$s_y=u_{y}t-\frac{1}{2}g{t}^2$$ $$v_y^2=u_y^2-2g{s}_y$$

where $s_y$ is the signed vertical displacement.

Vertical velocity changes uniformly because the acceleration is constant.

Same-Time Link

The most important connection:

Horizontal and vertical motions share the same time $t$.

This is the bridge that allows two separate component equations to be combined.

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Key Standard Results

These formulas apply to the common case:

• launch and landing at same vertical level
• air resistance neglected

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Time of Flight (Level Ground)

For a projectile that lands at the same vertical level as launch:

$$s_y=0$$

Using:

$$s_y=u_{y}t-\frac{1}{2}g{t}^2$$ $$0=(u\sin \theta )t-\frac{1}{2}g{t}^2$$

Solving gives:

$$t=0\text{or}T=\frac{2u\sin \theta }{g}$$

The non-zero solution $T$ is the time of flight.

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Maximum Height

At the highest point:

$$v_y=0$$

Using:

$$v_y^2=u_y^2-2g{s}_y$$

and

$$u_y=u\sin \theta$$

we get:

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

Therefore:

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

where $H$ is the maximum height above the launch point.

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Horizontal Range (Level Ground)

The horizontal range is:

$$R=u_{x}T$$

For a projectile that lands at the same vertical level as launch:

$$u_x=u\cos \theta ,T=\frac{2u\sin \theta }{g}$$

So:

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

This result is valid only when launch and landing heights are the same.

Maximum Range

For level launch and landing, with fixed launch speed $u$ and negligible air resistance:

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

Since:

$$\sin 2\theta \le 1$$

maximum range occurs when:

$$2\theta =90^{\circ }$$

so:

$$\theta =45^{\circ }$$

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Highest Point

At the top of the trajectory:

$$v_y=0$$

but:

$$a_y=-g$$

So the projectile is not in equilibrium.

Also:

$$v_x=u_x$$

still remains non-zero (unless launched vertically).

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Level-Ground Projectile Reasoning

For launch and landing at same height:

• ascent time = descent time
• path is symmetric (ideal model)
• speed at landing equals speed at launch
• launch angle = landing angle (same magnitude)

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Non-Level Launch / Landing (Qualitative)

If landing height differs from launch height:

• motion is no longer symmetric
• time up ≠ time down
• standard formulas for $T$, $R$, $H$ may need modification
• safest method: solve horizontal and vertical equations directly

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Worked Examples

Example 1: Component Resolution

A ball is launched at $20{\text{m s}}^{-1}$ at $30^{\circ }$ above the horizontal.

Resolving into components:

$$u_x=20\cos 30^{\circ }=17.3{\text{m s}}^{-1}$$ $$u_y=20\sin 30^{\circ }=10.0{\text{m s}}^{-1}$$

Taking upward as positive:

• initial horizontal velocity is $17.3{\text{m s}}^{-1}$
• initial vertical velocity is $10.0{\text{m s}}^{-1}$ upward

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Example 2: Time of Flight

A projectile is launched at $25{\text{m s}}^{-1}$ at $40^{\circ }$ above the horizontal.

For motion where launch and landing are at the same vertical level:

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

Substituting:

$$T=\frac{2(25)\sin 40^{\circ }}{9.81}=3.28\text{s}$$

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Example 3: Range

Using the same projectile ($u=25{\text{m s}}^{-1}$, $\theta =40^{\circ }$), and assuming launch and landing occur at the same vertical level:

$$R=\frac{u^2\sin 2\theta }{g}$$ $$R=\frac{25^2\sin 80^{\circ }}{9.81}=62.7\text{m}$$

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Use vertical motion to find time first, then horizontal motion:

$$s_x=u_{x}t$$

Example 4: Horizontal Launch from Table

A ball leaves a table horizontally at $6.0{\text{m s}}^{-1}$ from a height of $1.8\text{m}$.

Initial vertical velocity:

$$u_y=0$$

Take upward as positive:

$$s_y=-1.8,a_y=-g$$

Using vertical motion:

$$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$$ $$-1.8=-\frac{1}{2}g{t}^2$$ $$t=\sqrt{\frac{2(1.8)}{9.81}}=0.61\text{s}$$

Then horizontal motion:

$$s_x=u_{x}t=6.0\times 0.61=3.7\text{m}$$

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Common Exam Pitfalls

• mixing horizontal and vertical quantities in one equation
• forgetting to resolve initial velocity
• using different times for x and y motions
• assuming acceleration is zero at highest point
• using range formula when landing level differs
• using $u$ instead of $u_x$ horizontally
• wrong sign for vertical acceleration
• forgetting horizontal velocity is constant

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Exam Relevance

Projectile-motion questions test component resolution, sign convention, and the correct use of a common time variable across horizontal and vertical motion.

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Formula Summary

Components

$$u_x=u\cos \theta$$ $$u_y=u\sin \theta$$

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Horizontal

$$a_x=0$$ $$v_x=u_x$$ $$s_x=u_{x}t$$

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Vertical

$$a_y=-g$$ $$v_y=u_y-gt$$ $$s_y=u_{y}t-\frac{1}{2}g{t}^2$$

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Level Launch/Landing

$$T=\frac{2u\sin \theta }{g}$$ $$H=\frac{u^2{\sin }^2\theta }{2g}$$ $$R=\frac{u^2\sin 2\theta }{g}$$

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Links

Related Links

• Projectile Motion
• Kinematics
• Projectile and Relative Motion
• Constant Acceleration Models
• Vectors
• Forces
• Dynamics
• Circular Motion