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Engine 5 of 7

Vertical Motion Engine

Every Vertical Motion Problem is UAM Under Gravity

Purpose

Every vertical motion problem is simply Uniformly Accelerated Motion under gravity.

Students often memorize separate formulas for:

In reality, all of them are applications of one idea.

Acceleration Due to Gravity

5.1 Master Recognition Rule

Whenever you see any of the following trigger words:

Immediate Thought

Vertical Motion
↓
Gravity Present
↓
UAM Under Gravity
Golden Insight

All vertical motion formulas originate from:

$$a=\pm g$$

The sign depends on the chosen coordinate system.

5.2 Sign Convention Quick Box

Upward Positive

Quantity Sign
Upward Velocity +
Downward Velocity -
Gravity -g

Downward Positive

Quantity Sign
Downward Velocity +
Upward Velocity -
Gravity +g
Golden Rule

Choose any coordinate system.

Remain consistent throughout the problem.

5.3 Free Fall Engine

Recognition Trigger

Question says:

Immediate Observations

Initial Velocity

$$u=0$$
Acceleration

$$a=g$$

(with proper sign)

Master Formula Set

Use UAM formulas directly.

$$v=u+gt$$
$$s=ut+\frac12gt^2$$

Special Result

Since:

$$u=0$$

Therefore:

$$s=\frac12gt^2$$

Trigger Table

Question Says Immediate Action
Dropped u = 0
Released u = 0
Free Fall a = g
Falling Body Use UAM
Recognition Shortcut

Dropped
↓
Released from Rest
↓
$$u=0$$
↓
Apply UAM

5.4 Upward Projection Engine

Recognition Trigger

Immediate Setup

Usually choose:

Upward Positive
Quantity Sign
Initial Velocity +u
Gravity -g

Master Formula Set

$$v=u-gt$$
$$s=ut-\frac12gt^2$$
$$v^2=u^2-2gs$$

Recognition Shortcut

Thrown Upward
↓
UAM
↓
Replace
$$a$$
by
$$-g$$
Common Mistake

Students often memorize a separate chapter for upward projection.

It is simply UAM with:

$$a=-g$$

5.5 Highest Point Engine

One of the most important exam situations.

Recognition Trigger

Immediate Observation

At Highest Point

$$v=0$$
Golden Trap

Many students write:

$$a=0$$

Wrong.

Acceleration remains:

$$a=-g$$

Maximum Height

Using:

$$v^2=u^2-2gH$$

At highest point:

$$v=0$$

Therefore:

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

Time to Reach Highest Point

Using:

$$v=u-gt$$

At highest point:

$$v=0$$

Therefore:

$$t=\frac{u}{g}$$
Highest Point Summary

Velocity
↓
Zero

Acceleration
↓
Still Equal to Gravity

5.6 Return Journey Engine

Recognition Trigger

Immediate Thought

Symmetry Available

Time of Flight

Valid only when:

Launch Level
=
Landing Level
$$T=\frac{2u}{g}$$

Return Velocity

Magnitude:

Same as Initial Velocity

Direction:

Opposite

Key Results

Quantity Result
Net Displacement 0
Total Distance 2H
Return Speed Same as Launch Speed
Direction Opposite
Recognition Shortcut

Returns to Same Level
↓
Use Symmetry
↓
$$T=\frac{2u}{g}$$
Important Condition

This shortcut works only when launch and landing levels are identical.

5.7 Symmetry Engine

Recognition Trigger

Golden Symmetry Rule

At the Same Height

Speed Magnitude
=
Same

Example

During upward journey:

Speed = 15 m/s

At the same height during downward journey:

Speed = 15 m/s

Direction changes.

Magnitude remains the same.

Symmetry Shortcut

Same Height
↓
Same Speed Magnitude

Direction May Differ

5.8 Elevated Projection Engine

Recognition Trigger

Immediate Observation

Launch Level
≠
Landing Level
Most Important Trap

Many students use:

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

Wrong.

This formula requires:

Same Launch and Landing Levels

Correct Method

Use UAM directly.

Apply:

$$s=ut+\frac12at^2$$

using the actual height.

Recognition Shortcut

Tower Problem
↓
No Symmetry Shortcut
↓
Use Full UAM

5.9 Free Fall Shortcut Engine

Recognition Trigger

Successive Distance Ratio

For equal successive time intervals:

1 : 3 : 5 : 7 : 9 ...

Valid Only If

Common Trap

Do not use the ratio if:

Initial Velocity ≠ 0

5.10 Vertical Motion Decision Table

Trigger Seen Immediate Tool
Dropped Body u = 0
Free Fall UAM + g
Thrown Upward a = -g
Highest Point v = 0
Same Height Symmetry
Return to Launch Point Time of Flight
Tower Problem Elevated Projection
Successive Distances 1 : 3 : 5 : 7 ...

Top Exam Traps

Trap 1

Highest Point

$$v=0$$

but

$$a\neq0$$
Trap 2

Using

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

for Tower Problems
Trap 3

Wrong Sign of Gravity
Trap 4

Distance Confused with Displacement
Trap 5

Using Symmetry

when Heights are Different

One-Minute Revision Sheet

Dropped
↓
$$u=0$$

Thrown Upward
↓
$$a=-g$$

Highest Point
↓
$$v=0$$

Return Journey
↓
$$T=\frac{2u}{g}$$

Same Height
↓
Same Speed

Tower Problem
↓
No Symmetry Formula

Released from Rest
↓
1 : 3 : 5 : 7 ...
Golden Vertical Motion Rule

Every vertical motion problem is simply UAM with gravity.

Do not memorize separate chapters.

Recognize gravity, assign signs correctly, and apply the UAM toolkit.

Continue Revision

Sign Convention Engine

Review direction and gravity signs.

Relative Motion Engine

Learn observer-based motion analysis.

Graph Intelligence Engine

Decode motion using slope and area.

PYQ Masterclass

Practice exam-oriented applications.