Problem Solving: Projectiles
In August 2012, NASA's Curiosity rover entered Mars's atmosphere at 5.9 km/s during the 'Seven Minutes of Terror'. The entry arc was a multi-step projectile problem modified by drag: parachute deployed at 460 m/s, sky-crane lowered the rover to 2 m/s at touchdown. Six precision trajectory calculations — each a multi-step problem — had to succeed in sequence for the mission to survive.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
Write down every formula you know for projectile motion. Include what each variable means and its units.
Know
- The GIVEN → FIND → METHOD → ANSWER problem-solving protocol
- The six key projectile motion formulas and when each applies
- When the range equation $R = v^2\sin(2\theta)/g$ can and cannot be used
Understand
- Why launch height vs landing height determines which equations apply
- How to solve problems with partial information by combining vertical and horizontal equations
- Why time of flight depends on vertical motion only, not horizontal velocity
Can Do
- Apply the problem-solving protocol to multi-step projectile questions
- Identify which equation to use based on known and unknown quantities
- Solve projectile problems involving uneven ground or intermediate obstacles
Core Content
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.
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.
Problem-Solving Checklist
- Draw a diagram — label axes, launch angle, heights
- Resolve velocity: $v_x = v\cos\theta$, $v_y = v\sin\theta$
- Identify the direction of positive (usually up and to the right)
- Check launch vs landing height — same height → range equation OK
- Choose the right equation — match knowns/unknowns
- Substitute, calculate, check (units, sign, magnitude)
Use GIVEN → FIND → METHOD → ANSWER for every projectile problem. The critical check: if launch and landing heights differ, the range equation $R = v^2\sin(2\theta)/g$ does NOT apply — use the equations of motion instead.
Pause — copy the highlighted protocol into your book before moving on.
Did you get this? True or false: the range equation $R = v^2\sin(2\theta)/g$ can be used whenever a projectile is launched at an angle.
Using the range equation on level ground
We just saw the four-step protocol for tackling projectile problems. That raises a question: how does the protocol work in practice when the range equation can be used? This card answers it → worked example on level ground, rearranging $R = v^2\sin(2\theta)/g$ for launch speed.
A projectile is launched at 38° above the horizontal and lands 45 m away on level ground. Find the launch speed. Take $g = 9.8\ \text{m/s}^2$.
FIND: Launch speed $v$
To find launch speed from range on level ground, rearrange the range equation: $v = \sqrt{Rg/\sin(2\theta)}$. Always compute $\sin(2\theta)$ first, and confirm level-ground conditions before applying the formula.
Add the highlighted rearrangement to your notes before the check below.
Quick check: A ball is launched at 45° and lands 80 m away on level ground. Which value of $\sin(2\theta)$ is used in the range equation?
A complete analysis from launch to landing
We just saw how to find launch speed using the range equation. That raises a question: what if the question asks for maximum height, time, and range all together? This card answers it → use $h_{\text{max}} = v_y^2/(2g)$, $t_{\text{up}} = v_y/g$, $t_{\text{total}} = 2t_{\text{up}}$, and $R = v_x t_{\text{total}}$ in sequence.
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 $g = 9.8\ \text{m/s}^2$.
$v = 25\ \text{m/s}$, $\theta = 50°$, level ground, $g = 9.8\ \text{m/s}^2$
$v_x = 25\cos(50°) = 16.1\ \text{m/s}$, $\quad v_y = 25\sin(50°) = 19.2\ \text{m/s}$
For symmetrical flight on level ground: $h_{\text{max}} = v_y^2/(2g)$; $t_{\text{up}} = v_y/g$; $t_{\text{total}} = 2t_{\text{up}}$; range $R = v_x \times t_{\text{total}}$. Resolve components at the start before applying any formula.
Pause — copy the highlighted sequence of four formulas into your book before moving on.
Fill the gap: For symmetrical projectile flight on level ground, the total flight time equals times the time to reach maximum height.
When launch and landing heights differ
We just saw the standard sequence for level-ground problems. That raises a question: what happens when the launch and landing heights are different — like over a cliff or past a wall? This card answers it → heights differ, so find time from $t = s_x/v_x$, then compute $s_y$ and add reference height.
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?
Components: $v_x = 18\cos(35°) = 14.7\ \text{m/s}$, $v_y = 18\sin(35°) = 10.3\ \text{m/s}$
Step 4 — Clearance above wall:$$\text{Clearance} = 18.3 - 8 = \mathbf{10.3\ \text{m}}$$The ball clears the wall by approximately $\mathbf{10.3\ \text{m}}$.
When heights differ, the range equation does not apply. Find time from horizontal motion ($t = s_x/v_x$), compute vertical displacement ($s_y = v_y t - \tfrac{1}{2}g t^2$), add the reference height, then compare with any obstacle height.
Pause — write the highlighted uneven-ground procedure into your book.
Quick check: A stone is thrown from a cliff 30 m high. Why can’t you use $R = v^2\sin(2\theta)/g$ to find where it lands?
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 $v = 13\ \text{m/s}$, $\theta = 38°$, $h_0 = 2.2\ \text{m}$:
This is consistent with the world record, demonstrating how the projectile model applies to real athletic performance.
Activities · apply the protocol
Finding 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 $\theta$.
Projectile from a Balcony
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 $g = 9.8\ \text{m/s}^2$.
Minimum Speed to Clear a Tree
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.
Practise resolving and analysing motion
A projectile is launched at 18 m/s, 55° above the horizontal. Find the horizontal and vertical components of the initial velocity.
At a point in flight, the horizontal velocity is 12 m/s and the vertical velocity is 9 m/s. Find the resultant velocity (magnitude and direction).
A ball is thrown horizontally at 9 m/s from a cliff 39 m high. Calculate the time of flight and the horizontal distance travelled.
Expand to see book notes
Title: Problem Solving: Projectiles
The Problem-Solving Protocol:
- GIVEN — List all knowns with units and signs; resolve velocity into components
- FIND — State the unknown
- METHOD — Write the equation and principle
- ANSWER — Substitute, calculate, check
Key Formulas: $s_x = v\cos\theta \cdot t$ · $s_y = v\sin\theta \cdot t - \frac{1}{2}gt^2$ · $v_y = v\sin\theta - gt$ · $h_{\text{max}} = (v\sin\theta)^2/(2g)$ · $t_{\text{up}} = v\sin\theta/g$ · $R = v^2\sin(2\theta)/g$ (level ground only)
Critical Check: Is launch height = landing height? If not, the range equation does NOT apply. Use equations of motion instead.
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.
A fresh set drawn from this lesson’s question bank — feedback shown immediately.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Q1. 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. (3 marks)
Q2. 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)
Q3. 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? (4 marks)
Show all answers
Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Short Answer Model Answers
Q1 (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: Find time from horizontal motion, then vertical displacement.
$t = s_x / (v\cos\theta) = 4.0 / (8.5 \times \cos 52°) = 4.0 / 5.24 = 0.764\ \text{s}$
$s_y = 8.5\sin 52° \times 0.764 + \frac{1}{2}(-9.8)(0.764)^2 = 6.70 \times 0.764 - 2.86 = 5.12 - 2.86 = 2.26\ \text{m}$
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. The ball overshoots the basket by approximately 1.3 m.
Q2 (4 marks):
(a) Take down as negative: $s_y = -60\ \text{m}$, $v_y = 20\sin 40° = 12.86\ \text{m/s}$
$-60 = 12.86t - 4.9t^2 \Rightarrow 4.9t^2 - 12.86t - 60 = 0$
$t = (12.86 + \sqrt{12.86^2 + 4(4.9)(60)}) / 9.8 = (12.86 + 36.6) / 9.8 = 5.05\ \text{s}$ (rejecting negative root)
(b) $v_x = 20\cos 40° = 15.32\ \text{m/s}$; $s_x = 15.32 \times 5.05 = 77.4\ \text{m}$
(c) $v_y = 12.86 - 9.8(5.05) = -36.6\ \text{m/s}$; $v = \sqrt{15.32^2 + 36.6^2} = \sqrt{234.7 + 1339.6} = \sqrt{1574} = 39.7\ \text{m/s}$
Q3 (4 marks):
Using the range equation first is effective only when: (i) launch and landing heights are the same; (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; (iii) there is an intermediate obstacle (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.
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering projectile motion questions. Lighter alternative to the boss.
At the start you were told about NASA's Mars Science Laboratory 2012 entry — 5.9 km/s entry speed, 6 sequential trajectory calculations in 7 minutes, each one a multi-step projectile problem.
The GIVEN→FIND→METHOD→ANSWER protocol is precisely what NASA engineers use in a more rigorous form. Each stage of the MSL descent was solved by isolating the known quantities (entry speed, altitude, drag coefficients), identifying the target variable (parachute deployment speed at 460 m/s, final lowering speed at 2 m/s), choosing the right kinematic equation, then calculating. The method scales from a textbook question to a planetary landing — the structure is the same.