Learn

Questions

Game

Apply the problem-solving protocol to projectile questions

Solve problems with partial information

Identify which equation to use based on known and unknown quantities

Think First

RECALL

Write down every formula you know for projectile motion. Include what each variable means and its units.

Autosaved

Year 12 Physics Module 5: Advanced Mechanics 45 min Lesson 3 of 18 IQ1: Projectile Motion

Problem Solving: Projectiles

Apply the GIVEN→FIND→METHOD→ANSWER protocol to multi-step projectile problems. Work through problems involving partial information, symmetrical flight, and projectiles over uneven ground.

🎯

The Problem-Solving Protocol

A structured approach to every projectile question

Projectile Worked Example

Problem Solving Projectiles

Multi-step projectile problems can seem overwhelming. The key is to follow a consistent protocol that breaks the problem into manageable steps. Every worked example in this lesson follows the same format.

GIVEN → FIND → METHOD → ANSWER

GIVEN: List all known quantities with units and signs. Resolve the launch velocity into horizontal and vertical components immediately: $v_x = v \cos\theta$ and $v_y = v \sin\theta$.

FIND: State clearly what you need to calculate.

METHOD: Write the equation(s) you will use, rearranged for the unknown. Name the principle (e.g., “equation of motion in vertical direction”).

ANSWER: Substitute values with units, calculate, and check reasonableness.

Before calculating anything, always ask yourself: Is the launch height the same as the landing height? If not, the range equation $R = v^2 \sin(2\theta)/g$ does not apply.

f(x)

Problem-Solving Checklist

1. Draw a diagram

Label axes, launch angle, heights

2. Resolve velocity

$v_x = v\cos\theta$, $v_y = v\sin\theta$

3. Identify the direction of +

Usually up and to the right

4. Check launch vs landing height

Same height → range equation OK

5. Choose the right equation

Match knowns/unknowns

6. Substitute, calculate, check

Units, sign, magnitude

Worked Example 1: Finding Launch Speed from Range

Using the range equation on level ground

Worked Example 1

A projectile is launched at 38° above the horizontal and lands 45 m away on level ground. Find the launch speed. Take $g = 9.8\ \text{m/s}^2$.

GIVEN

$R = 45\ \text{m}$, $\theta = 38°$, level ground, $g = 9.8\ \text{m/s}^2$

FIND

Launch speed $v$

METHOD

Use the range equation for level ground:

$$R = \frac{v^2 \sin(2\theta)}{g} \quad \Rightarrow \quad v = \sqrt{\frac{Rg}{\sin(2\theta)}}$$

ANSWER

$$v = \sqrt{\frac{45 \times 9.8}{\sin(76°)}} = \sqrt{\frac{441}{0.970}} = \sqrt{454} = 21.3\ \text{m/s}$$

The launch speed is $21.3\ \text{m/s}$ (3 s.f.).

Check: $\sin(2 \times 38°) = \sin(76°) \approx 0.970$. The answer is reasonable for a projectile travelling 45 m.

Worked Example 2: Maximum Height, Time and Range

A complete analysis from launch to landing

Worked Example 2

A ball is thrown at $25\ \text{m/s}$ at $50°$ above the horizontal from ground level. Find: (a) the maximum height, (b) the time to reach maximum height, (c) the total flight time, and (d) the horizontal range. Take $g = 9.8\ \text{m/s}^2$.

GIVEN

$v = 25\ \text{m/s}$, $\theta = 50°$, $g = 9.8\ \text{m/s}^2$, level ground

Components: $v_x = 25\cos(50°) = 16.1\ \text{m/s}$, $v_y = 25\sin(50°) = 19.2\ \text{m/s}$

FIND

(a) $h_{\text{max}}$, (b) $t_{\text{up}}$, (c) $t_{\text{total}}$, (d) $R$

METHOD

Separate into vertical and horizontal. For (a) and (b), use vertical motion. At max height, $v_y = 0$.

ANSWER

(a) Maximum height:

$$v_y^2 = u_y^2 + 2as \Rightarrow 0 = (25\sin 50°)^2 - 2(9.8)h_{\text{max}}$$

$$h_{\text{max}} = \frac{(25\sin 50°)^2}{2 \times 9.8} = \frac{19.15^2}{19.6} = \frac{366.7}{19.6} = 18.7\ \text{m}$$

(b) Time to maximum height:

$$v_y = u_y - gt \Rightarrow 0 = 25\sin 50° - 9.8t_{\text{up}}$$

$$t_{\text{up}} = \frac{25\sin 50°}{9.8} = \frac{19.15}{9.8} = 1.95\ \text{s}$$

(c) Total flight time (symmetrical flight):

$$t_{\text{total}} = 2 \times t_{\text{up}} = 2 \times 1.95 = 3.91\ \text{s}$$

(d) Horizontal range:

$$R = v_x \times t_{\text{total}} = 25\cos 50° \times 3.91 = 16.07 \times 3.91 = 62.8\ \text{m}$$

Worked Example 3: Projectile Over a Wall

When launch and landing heights differ

Worked Example 3

A ball is kicked at $18\ \text{m/s}$ at $35°$ above horizontal from the edge of a cliff that is 15 m above sea level. A wall stands 25 m horizontally from the cliff edge and is 8 m high (measured from sea level). Does the ball clear the wall? If so, by how much?

GIVEN

$v = 18\ \text{m/s}$, $\theta = 35°$, cliff height $= 15\ \text{m}$, wall distance $s_x = 25\ \text{m}$, wall height $= 8\ \text{m}$, $g = 9.8\ \text{m/s}^2$

Components: $v_x = 18\cos(35°) = 14.7\ \text{m/s}$, $v_y = 18\sin(35°) = 10.3\ \text{m/s}$

FIND

Does the ball clear the 8 m wall? If yes, by what vertical margin?

METHOD

1. Find time to reach wall using horizontal motion ($s_x = v_x t$).
2. Find vertical displacement above launch at that time using $s_y = v_y t + \frac{1}{2}(-g)t^2$.
3. Add cliff height to get height above sea level.
4. Compare with wall height.

ANSWER

Step 1 — Time to reach wall:

$$t = \frac{s_x}{v_x} = \frac{25}{18\cos 35°} = \frac{25}{14.74} = 1.70\ \text{s}$$

Step 2 — Vertical displacement at $t = 1.70$ s:

$$s_y = 18\sin 35° \times 1.70 + \frac{1}{2}(-9.8)(1.70)^2$$

$$s_y = 10.32 \times 1.70 - 4.9 \times 2.89 = 17.5 - 14.2 = 3.3\ \text{m}$$

The ball is 3.3 m above the cliff edge when it reaches the wall.

Step 3 — Height above sea level:

$$\text{Height} = 15 + 3.3 = 18.3\ \text{m}$$

Step 4 — Clearance above wall:

$$\text{Clearance} = 18.3 - 8 = 10.3\ \text{m}$$

The ball clears the wall by approximately $10.3\ \text{m}$. Note: small differences due to rounding in intermediate steps are acceptable. Using more precise values gives $10.4\ \text{m}$.

f(x)

Key Formulas — Projectile Motion

$s_x = v\cos\theta \cdot t$

Horizontal displacement

$s_y = v\sin\theta \cdot t + \frac{1}{2}(-g)t^2$

Vertical displacement

$v_y = v\sin\theta - gt$

Vertical velocity at time $t$

$h_{\text{max}} = \frac{(v\sin\theta)^2}{2g}$

Maximum height (above launch)

$t_{\text{up}} = \frac{v\sin\theta}{g}$

Time to maximum height

$R = \frac{v^2\sin(2\theta)}{g}$

Range on level ground only

Common Misconceptions

✗

“Use the range equation for all problems”

Right: The range equation $R = v^2\sin(2\theta)/g$ only applies on level ground. Always check if launch and landing heights differ. If they do, solve using the equations of motion separately.

✗

“Time of flight depends on horizontal velocity”

Right: Time is determined entirely by vertical motion. The horizontal velocity $v_x$ affects range but not flight time (ignoring air resistance).

✗

“You need to know the mass of the projectile”

Right: Projectile motion in a vacuum (or neglecting air resistance) is completely independent of mass. All objects follow the same trajectory for the same launch conditions.

Real World: Olympic Shot Put

The shot put is launched from approximately 2.2 m height (the athlete's shoulder) at roughly $13\ \text{m/s}$ and $38°$. The optimal angle is less than 45° because the launch is already elevated — a lower angle gives more horizontal speed while still allowing sufficient air time from the raised position.

The current men's world record is 23.56 m (Randy Barnes, 1990). Using the elevated launch model with $v = 13\ \text{m/s}$, $\theta = 38°$, $h_0 = 2.2\ \text{m}$:

$$s_y = -2.2 = 13\sin(38°) \cdot t - 4.9t^2 \Rightarrow t \approx 2.1\ \text{s}, \quad R \approx 24\ \text{m}$$

This is consistent with the world record, demonstrating how the projectile model applies to real athletic performance.

Activities

Apply the protocol to these problems

Activity 1: Find the Launch Angle

A javelin is thrown at $28\ \text{m/s}$ and lands 68 m away on level ground. Use the GIVEN→FIND→METHOD→ANSWER protocol to find the launch angle.

Hint: Rearrange $R = v^2\sin(2\theta)/g$ for $\theta$.

Activity 2: Projectile from a Height

A ball is thrown from a balcony 12 m above the ground at $15\ \text{m/s}$, $40°$ above horizontal. Find:

(a) the time to hit the ground
(b) the horizontal distance travelled
(c) the speed at impact

Use $g = 9.8\ \text{m/s}^2$.

Activity 3: Clearing an Obstacle

A golfer wants to clear a tree that is 30 m away horizontally and 15 m high. The ball is struck from ground level. What is the minimum launch speed required at $45°$?

Hint: The ball must be at least 15 m high when $s_x = 30$ m. Set up simultaneous equations for horizontal and vertical motion.

Copy into Books

Title: Problem Solving: Projectiles

The Problem-Solving Protocol:

GIVEN — List all knowns with units and signs
FIND — State the unknown
METHOD — Write the equation and principle
ANSWER — Substitute, calculate, check

Key Formulas:

$s_x = v\cos\theta \cdot t$
$s_y = v\sin\theta \cdot t + \frac{1}{2}(-g)t^2$
$v_y = v\sin\theta - gt$
$h_{\text{max}} = (v\sin\theta)^2/(2g)$
$t_{\text{up}} = v\sin\theta/g$
$R = v^2\sin(2\theta)/g$ (level ground only)

Critical Check: Is launch height = landing height? If not, the range equation does NOT apply. Use equations of motion instead.

Worked Example Summary: We 3 problems: finding launch speed from range, complete trajectory analysis, and projectile over a wall (uneven ground).

Revisit: Think First

Look back at the formulas you wrote at the start. Did you include all six key equations? Are your variable definitions and units correct? Add any you missed.

Interactive: Projectile Solver Interactive

Questions

Key Terms

Range ($R$) Horizontal distance from launch to landing

Time of flight Total elapsed time from launch to landing

Maximum height Peak vertical position above the launch point

Components $v_x = v\cos\theta$, $v_y = v\sin\theta$

Symmetrical flight Launch and landing at the same height

Problem-solving protocol GIVEN → FIND → METHOD → ANSWER

A projectile lands 5 m below its launch point. Which equation correctly finds flight time?

A $t = \frac{2v\sin\theta}{g}$

B Solve $-5 = v\sin\theta \cdot t - \frac{1}{2}gt^2$ for $t$

C $t = \frac{R}{v\cos\theta}$, then use that in the vertical equation

D $t = \sqrt{\frac{2 \times 5}{g}}$

Correct: B — When landing below launch, the vertical displacement is negative. You must use the full equation of motion $s_y = v\sin\theta \cdot t + \frac{1}{2}(-g)t^2$ and solve the quadratic for $t$.

A uses the level-ground formula which assumes the projectile returns to the same height. C uses circular logic — you don't know $R$ yet. D treats the projectile as if it were simply dropped, ignoring the initial upward velocity.

A ball is thrown at $30°$ and lands 40 m away on level ground. What was the launch speed? Take $g = 9.8\ \text{m/s}^2$.

A $15.3\ \text{m/s}$

B $21.3\ \text{m/s}$

C $30.0\ \text{m/s}$

D $12.7\ \text{m/s}$

Correct: B — $v = \sqrt{Rg/\sin(2\theta)} = \sqrt{40 \times 9.8 / \sin(60°)} = \sqrt{392/0.866} = \sqrt{453} = 21.3\ \text{m/s}$.

A uses $\sin\theta$ instead of $\sin(2\theta)$. C is a guess equalling the angle. D uses $\theta = 60°$ instead of $2\theta = 60°$ in the wrong way.

At the top of a projectile's path:

A The acceleration is momentarily zero

B Only horizontal acceleration acts

C The velocity is purely horizontal

D The velocity is zero

Correct: C — At the peak, the vertical component of velocity is zero, but the horizontal component remains $v_x = v\cos\theta$. Gravity still acts with $a = g$ downward.

A confuses zero vertical velocity with zero acceleration. B invents a horizontal acceleration that doesn't exist (air resistance is neglected). D confuses vertical velocity with total velocity.

A stone is thrown from a cliff at $15\ \text{m/s}$, $25°$ above horizontal. It hits the sea 30 m below. What determines the flight time?

A Horizontal velocity

B Vertical component of velocity and height

C Horizontal distance to the water

D Launch speed only

Correct: B — Flight time is determined entirely by vertical motion. The initial vertical velocity ($v_y = 15\sin 25°$) and the vertical displacement ($-30$ m) together determine $t$ through the equation $s_y = v_y t + \frac{1}{2}(-g)t^2$.

A and C incorrectly link time to horizontal quantities. D ignores both the launch angle (which sets $v_y$) and the height of the cliff.

The GIVEN→FIND→METHOD→ANSWER protocol requires you to:

A Memorise every formula

B List knowns, state unknowns, choose equation, then calculate

C Start calculating immediately

D Draw a diagram only

Correct: B — The protocol is a structured problem-solving process: list what you know (GIVEN), state what you need (FIND), select the right equation (METHOD), then substitute and solve (ANSWER).

A is ineffective — physics rewards understanding, not rote memorisation. C leads to errors because you haven't planned. D is incomplete — a diagram helps but doesn't replace the full protocol.

Short Answer Questions

3 marks Apply Band 4-5

A basketball player shoots at $8.5\ \text{m/s}$, $52°$ from horizontal from a height of $2.1\ \text{m}$. The basket is $3.05\ \text{m}$ high and $4.0\ \text{m}$ away horizontally. Determine whether the ball goes through the basket. Show all working using the problem-solving protocol.

4 marks Analyse Band 5-6

A stone is thrown from the top of a $60\ \text{m}$ cliff at $20\ \text{m/s}$, $40°$ above horizontal. Calculate:

(a) the time to reach the water
(b) the horizontal distance from the cliff base where it lands
(c) the speed at impact

Take $g = 9.8\ \text{m/s}^2$.

4 marks Evaluate Band 5-6

Evaluate the problem-solving strategy of “always using the range equation first” when approaching projectile motion questions. Under what circumstances is this effective, and when does it lead to errors?

View Model Answers

Short Answer 1 (3 marks)

GIVEN: $v = 8.5\ \text{m/s}$, $\theta = 52°$, $h_{\text{launch}} = 2.1\ \text{m}$, $h_{\text{basket}} = 3.05\ \text{m}$, $s_x = 4.0\ \text{m}$

FIND: Vertical position of the ball when $s_x = 4.0\ \text{m}$

METHOD: First find time from horizontal motion, then find vertical displacement.

$$t = \frac{s_x}{v\cos\theta} = \frac{4.0}{8.5 \times \cos 52°} = \frac{4.0}{5.24} = 0.764\ \text{s}$$

$$s_y = 8.5\sin 52° \times 0.764 + \frac{1}{2}(-9.8)(0.764)^2$$

$$s_y = 6.70 \times 0.764 - 2.86 = 5.12 - 2.86 = 2.26\ \text{m}$$

Total height above ground: $2.1 + 2.26 = 4.36\ \text{m}$

The basket is at $3.05\ \text{m}$. The ball is $4.36 - 3.05 = 1.31\ \text{m}$ above the basket.

Conclusion: The ball overshoots the basket by approximately 1.3 m.

Short Answer 2 (4 marks)

(a) Time to reach the water

Take down as negative: $s_y = -60\ \text{m}$, $v_y = 20\sin 40° = 12.86\ \text{m/s}$ (up, positive)

$$s_y = v_y t + \frac{1}{2}(-g)t^2 \Rightarrow -60 = 12.86t - 4.9t^2$$

$$4.9t^2 - 12.86t - 60 = 0$$

$$t = \frac{12.86 + \sqrt{12.86^2 + 4(4.9)(60)}}{9.8} = \frac{12.86 + \sqrt{165.3 + 1176}}{9.8} = \frac{12.86 + 36.6}{9.8} = 5.05\ \text{s}$$

(Rejecting the negative root $t = -2.43$ s)

(b) Horizontal distance

$$v_x = 20\cos 40° = 15.32\ \text{m/s}$$

$$s_x = 15.32 \times 5.05 = 77.4\ \text{m}$$

(c) Speed at impact

$$v_x = 15.32\ \text{m/s} \quad \text{(constant)}$$

$$v_y = 12.86 - 9.8(5.05) = 12.86 - 49.5 = -36.6\ \text{m/s}$$

$$v = \sqrt{15.32^2 + (-36.6)^2} = \sqrt{234.7 + 1339.6} = \sqrt{1574} = 39.7\ \text{m/s}$$

Short Answer 3 (4 marks)

Using the range equation first is effective only under specific conditions:

(i) Launch and landing heights are the same (level ground)
(ii) The question asks for range, launch speed, or launch angle
(iii) Air resistance is negligible

It leads to errors when:

(i) Launch and landing heights differ (most real-world problems)
(ii) The question asks for maximum height or time of flight specifically
(iii) There is an intermediate obstacle to check (e.g., does it clear a wall?)

Better strategy: Always resolve into components first, then use the appropriate equation of motion for the specific unknown. This works for all projectile problems, not just the special case of level ground. The range equation is a time-saver for a subset of problems, not a universal starting point.

Boss Battle

I have completed this lesson