Kinematics: Projectiles

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your WAEC exam.

Projectile Motion — Key Facts

A projectile is any object launched into the air that is acted upon only by gravity (and air resistance, which we neglect at WAEC level). The path traced is called the trajectory, and it is always parabolic.

Essential Formulas

• Horizontal velocity: remains constant — $u_x = u \cos \theta$
• Vertical velocity: changes due to gravity — $u_y = u \sin \theta$
• Time of flight: $T = \dfrac{2u \sin \theta}{g}$
• Maximum height: $H = \dfrac{u^2 \sin^2 \theta}{2g}$
• Horizontal range: $R = \dfrac{u^2 \sin 2\theta}{g}$

Key Facts to Memorise

• At maximum height, vertical velocity = 0
• Time to reach max height = $\dfrac{u \sin \theta}{g}$ (half of total flight time)
• For $\theta = 45°$, range $R$ is maximum: $R_{\max} = \dfrac{u^2}{g}$
• The horizontal and vertical motions are independent — resolve everything at the start
• Acceleration due to gravity, $g = 9.8 , \text{m s}^{-2}$ (use $10 , \text{m s}^{-2}$ in exams for simplicity)

⚡ WAEC Exam Tip: The most common mistake is mixing up horizontal and vertical components. Always resolve the initial velocity $u$ at the start. If a question gives speed and angle, find $u_x$ and $u_y$ first before doing anything else. WAEC often asks: “Find the time of flight” or “Find the range” — these formulas are directly examinable.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for building a solid understanding of projectile motion.

Understanding Projectile Motion

Projectile motion is a cornerstone of WAEC Physics. It combines two straight-line motions — horizontal (constant velocity) and vertical (uniform acceleration) — into one elegant framework. The object moves freely through the air after being projected, with only gravitational force acting upon it.

The Two-Component Approach

Consider an object projected with initial speed $u$ at an angle $\theta$ to the horizontal:

Component	Velocity	Acceleration	Displacement
Horizontal	$u_x = u \cos \theta$ (constant)	$a_x = 0$	$x = u \cos \theta \cdot t$
Vertical	$u_y = u \sin \theta$	$a_y = -g$	$y = u \sin \theta \cdot t - \frac{1}{2}gt^2$

The parabolic trajectory arises because horizontal displacement grows linearly with time while vertical displacement follows a quadratic relationship.

Deriving the Key Quantities

Time of Flight: The projectile returns to the ground when $y = 0$. Using $y = u \sin \theta \cdot t - \frac{1}{2}gt^2$ and setting $y = 0$:

$$t(u \sin \theta - \frac{1}{2}gt) = 0 \implies t = 0 \quad \text{or} \quad t = \frac{2u \sin \theta}{g}$$

The non-zero solution gives the total time of flight: $T = \frac{2u \sin \theta}{g}$.

Maximum Height: At the peak, vertical velocity becomes zero: $v_y = u \sin \theta - gt = 0$, giving $t_{\text{peak}} = \frac{u \sin \theta}{g}$. Substituting into the displacement equation:

$$H = (u \sin \theta)\left(\frac{u \sin \theta}{g}\right) - \frac{1}{2}g\left(\frac{u \sin \theta}{g}\right)^2 = \frac{u^2 \sin^2 \theta}{2g}$$

Horizontal Range: $R = u_x \times T = (u \cos \theta) \times \left(\frac{2u \sin \theta}{g}\right) = \frac{u^2 \sin 2\theta}{g}$.

Worked Example (WAEC Style)

A ball is kicked with a velocity of $40 , \text{m s}^{-1}$ at $30°$ to the horizontal. Calculate: (a) the time of flight (b) the maximum height reached (c) the horizontal range

Solution: Given: $u = 40 , \text{m s}^{-1}$, $\theta = 30°$, $g = 10 , \text{m s}^{-2}$

(a) Time of flight: $T = \frac{2u \sin \theta}{g} = \frac{2 \times 40 \times \sin 30°}{10} = \frac{80 \times 0.5}{10} = 4.0 , \text{s}$

(b) Max height: $H = \frac{u^2 \sin^2 \theta}{2g} = \frac{40^2 \times (0.5)^2}{2 \times 10} = \frac{1600 \times 0.25}{20} = 20 , \text{m}$

(c) Range: $R = \frac{u^2 \sin 2\theta}{g} = \frac{40^2 \times \sin 60°}{10} = \frac{1600 \times 0.866}{10} = 138.6 , \text{m}$ (or $138.6 , \text{m}$)

⚡ WAEC Exam Tip: Always state the formula before substituting numbers. Examiners award marks for correct formula application even if arithmetic is slightly off. Also note: if the question says “neglect air resistance,” use the standard equations as above. If it gives you a launch speed of $v$ with no angle, you cannot solve for range or time unless an angle is given or implied.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage including derivations and WAEC past question patterns.

Full Derivation: Equation of the Trajectory

Eliminating time $t$ from the two displacement equations gives the Cartesian equation of the trajectory. From horizontal displacement: $t = \frac{x}{u \cos \theta}$. Substituting into vertical displacement:

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

Since $\frac{1}{\cos^2 \theta} = \sec^2 \theta = 1 + \tan^2 \theta$, this can be rewritten as a parabola in standard form, confirming that all projectile trajectories under uniform gravity are parabolic.

The Independence Principle

Galileo first established that horizontal and vertical motions are independent. This means:

• A ball dropped from height $h$ and another fired horizontally from the same height will hit the ground at the same time
• The horizontal throw has no effect on vertical fall time

This principle simplifies problem-solving significantly — solve each direction separately, linked only by the common variable time.

Projection from a Height (NOT level ground)

If a projectile is launched from a height $h$ above the ground (rather than from ground level), the equation $y = u \sin \theta \cdot t - \frac{1}{2}gt^2 + h$ applies. The time to reach the ground is longer than for a level-ground projection at the same speed and angle.

For this case, set $y = 0$ and solve the quadratic: $-\frac{1}{2}gt^2 + u \sin \theta \cdot t + h = 0$, giving:

$$t = \frac{u \sin \theta + \sqrt{u^2 \sin^2 \theta + 2gh}}{g}$$

WAEC Past Question Patterns

From WAEC Physics papers (2020–2024), projectile questions typically:

1. Give initial speed $u$ and angle $\theta$ — ask for $T$, $H$, or $R$ (direct formula substitution)
2. Ask which quantity remains constant (horizontal velocity) — multiple choice
3. Describe a footballer or cricketer — apply range/time formulas to a sporting context
4. Ask about the velocity at the maximum height (only horizontal component remains, so $v = u \cos \theta$)

⚡ Common Pitfalls to Avoid

• Using $g = 9.8$ when the question expects $g = 10$ — check the paper’s convention
• Forgetting that $\sin 2\theta = 2\sin\theta\cos\theta$ and using wrong double-angle values
• Mixing up “time to peak” with “time of flight” — peak is always half of total flight for level ground
• Writing units incorrectly — speeds need $\text{m s}^{-1}$, ranges need metres

Key Derivations to Revise

1. $T = \frac{2u \sin \theta}{g}$ from $y = u \sin \theta \cdot t - \frac{1}{2}gt^2 = 0$
2. $H = \frac{u^2 \sin^2 \theta}{2g}$ by substituting $t_{\text{peak}} = \frac{u \sin \theta}{g}$ into vertical displacement
3. $R = \frac{u^2 \sin 2\theta}{g}$ by combining $T$ with horizontal displacement
4. Trajectory equation $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$

Quick Reference Table

Quantity	Formula
Horizontal velocity	$u_x = u \cos \theta$
Vertical velocity	$u_y = u \sin \theta$
Time of flight	$T = \frac{2u \sin \theta}{g}$
Maximum height	$H = \frac{u^2 \sin^2 \theta}{2g}$
Horizontal range	$R = \frac{u^2 \sin 2\theta}{g}$
Max range angle	$\theta = 45°$
Velocity at max height	$v = u \cos \theta$ (horizontal only)

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