Projectile motion

General physics

🎯 Learning Objectives

Describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction.

🗣️ Language Objectives

Define and correctly use terms such as projectile, trajectory, range, and initial velocity.
Describe the horizontal and vertical components of projectile motion.
Explain the reasoning behind the solutions to projectile motion problems.

Translation (Kazakh)

Снаряд, траектория, қашықтық және бастапқы жылдамдық сияқты терминдерді анықтау және дұрыс қолдану.
Снаряд қозғалысының горизонталь және вертикаль құрамдас бөліктерін сипаттау.
Снаряд қозғалысы есептерінің шешімдерінің негіздемесін түсіндіру.

🔑 Key Terms

English	Русский	Қазақша
Projectile Motion	Движение снаряда	Снаряд қозғалысы
Trajectory	Траектория	Траектория
Range	Дальность полета	Ұшу қашықтығы
Initial Velocity	Начальная скорость	Бастапқы жылдамдық
Horizontal Component	Горизонтальная составляющая	Горизонталь құраушы
Vertical Component	Вертикальная составляющая	Вертикаль құраушы

🃏 Flashcards

Glossary

Projectile: An object that is thrown or projected into the air, subject only to the acceleration of gravity.

Translation

Ауаға лақтырылған немесе атылған, тек ауырлық күшінің үдеуіне бағынатын нысан.

Trajectory: The path followed by a projectile.

Translation

Снарядтың жүріп өткен жолы.

Range: The horizontal distance traveled by a projectile.

Translation

Снарядтың горизонталь бағытта жүріп өткен қашықтығы.

📖 Theory

Projectile Motion

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The key to analyzing projectile motion is to treat the horizontal and vertical components of the motion -separately-. This is because the force of gravity only acts -vertically-, meaning it only affects the vertical motion. There are no horizontal forces (ignoring air resistance), so the horizontal velocity is constant.

Horizontal Motion

The horizontal component of velocity, v_x, is constant throughout the flight. The horizontal displacement, x, can be found using:

x = v_xt

Vertical Motion

The vertical component of velocity, v_y, changes due to gravity. The vertical motion is described by the standard equations for uniformly accelerated motion, with the acceleration being g (approximately 9.81 m/s² downwards).

v_y = u_y + gt
y = u_yt + ½gt²
v_y² = u_y² + 2gy

Here, u_y is the initial vertical velocity. The direction (up or down) is crucial, so we often take ‘up’ as positive, making g = -9.81 m/s².

By -combining- these two independent motions, we can fully describe the parabolic trajectory of a projectile.

Translation (Kazakh)

Снаряд қозғалысы

Снаряд қозғалысы дегеніміз — ауаға лақтырылған немесе атылған, тек ауырлық күшінің үдеуіне бағынатын нысанның қозғалысы. Снаряд қозғалысын талдаудың кілті — қозғалыстың горизонталь және вертикаль құрамдас бөліктерін бөлек қарастыру. Себебі ауырлық күші тек вертикаль бағытта әсер етеді, яғни ол тек вертикаль қозғалысқа әсер етеді. Горизонталь күштер жоқ (ауа кедергісін ескермегенде), сондықтан горизонталь жылдамдық тұрақты.

Горизонталь қозғалыс

Жылдамдықтың горизонталь құраушысы, v_x, ұшу бойы тұрақты болады. Горизонталь орын ауыстыруды, x, келесі формуламен табуға болады:

x = v_xt

Вертикаль қозғалыс

Жылдамдықтың вертикаль құраушысы, v_y, ауырлық күшінің әсерінен өзгереді. Вертикаль қозғалыс бірқалыпты үдемелі қозғалыстың стандартты теңдеулерімен сипатталады, мұндағы үдеу g (шамамен 9.81 м/с² төмен қарай).

v_y = u_y + gt
y = u_yt + ½gt²
v_y² = u_y² + 2gy

Мұндағы, u_y — бастапқы вертикаль жылдамдық. Бағыт (жоғары немесе төмен) маңызды, сондықтан біз жиі ‘жоғарыны’ оң деп аламыз, бұл g = -9.81 м/с² етеді.

Осы екі тәуелсіз қозғалысты біріктіре отырып, біз снарядтың параболалық траекториясын толық сипаттай аламыз.

Questions on the Theory

(Easy) What is the acceleration of a projectile in the horizontal direction, assuming no air resistance?
(Medium) Why is it useful to resolve the motion of a projectile into horizontal and vertical components?
(Medium) A ball is thrown horizontally from a cliff. At the same instant, an identical ball is dropped vertically from the same height. Which ball hits the ground first? Explain your reasoning.
(Hard) How would the trajectory of a projectile on the Moon, where the acceleration due to gravity is about 1/6th of that on Earth, differ from its trajectory on Earth, assuming the same initial velocity? Explain the differences in terms of range and maximum height.

Answer

The acceleration in the horizontal direction is zero.
Because the two components are independent of each other. The horizontal motion has constant velocity, while the vertical motion has constant acceleration (due to gravity). This simplifies the analysis and calculations.
They both hit the ground at the same time. The vertical motion determines the time of flight. Since both balls start with zero vertical velocity and fall the same vertical distance under the same acceleration (g), their time of flight will be identical.
On the Moon, the projectile would travel much farther (greater range) and reach a much higher peak. The constant horizontal velocity remains the same, but the time of flight would be significantly longer because the downward vertical acceleration is much smaller. A smaller ‘g’ means it takes more time for the vertical velocity to reverse and for the projectile to return to its initial height. The maximum height (where v_y = 0) would also be greater because the initial vertical velocity is counteracted by a smaller acceleration, allowing it to travel higher before stopping.

🧠 Exercises on Memorization

Match the term with its description:

Trajectory
Range
Horizontal Component
Vertical Component

A. The motion component affected by gravity.

B. The path a projectile follows through the air.

C. The motion component with constant velocity.

D. The total horizontal distance covered by a projectile.

Answer

1 — B

2 — D

3 — C

4 — A

▶️ Video

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📝 Solved Examples

Problem 1Problem 2

A ball is kicked from the ground with an initial velocity of 25 m/s at an angle of 30° to the horizontal. Find the time of flight and the range of the projectile.

Solution

First, resolve the initial velocity into horizontal and vertical components:

Initial horizontal velocity (u_x) = 25 * cos(30°) = 21.65 m/s
Initial vertical velocity (u_y) = 25 * sin(30°) = 12.5 m/s

To find the time of flight:
Consider the vertical motion. The time of flight is the time it takes for the ball to return to the ground (y=0).
Using y = u_yt + ½gt²
0 = 12.5t + ½(-9.81)t²
0 = t(12.5 — 4.905t)
One solution is t=0 (the start), the other is:
12.5 = 4.905t
t = 12.5 / 4.905 = 2.55 s
So, the time of flight is 2.55 s.

To find the range:
Use the horizontal motion equation.
Range (x) = u_x * t
x = 21.65 m/s * 2.55 s
x = 55.2 m
So, the range is 55.2 m.

A stone is thrown horizontally with a speed of 15 m/s from the top of a cliff 75 m high. How far from the base of the cliff does the stone land?

Solution

First, analyze the vertical and horizontal motions separately.

Vertical motion:

Initial vertical velocity (u_y) = 0 m/s
Vertical displacement (y) = -75 m (negative as it’s downwards)
Acceleration (g) = -9.81 m/s²

We need to find the time (t) it takes to fall.
Using y = u_yt + ½gt²
-75 = (0)t + ½(-9.81)t²
-75 = -4.905t²
t² = -75 / -4.905 = 15.29
t = √15.29 = 3.91 s

Horizontal motion:

Horizontal velocity (v_x) = 15 m/s (constant)
Time (t) = 3.91 s

Now, calculate the range (horizontal distance, x).
x = v_x * t
x = 15 m/s * 3.91 s
x = 58.65 m
The stone lands 58.7 m from the base of the cliff.

🔬 Investigation

Use the PhET simulation to explore how different factors affect a projectile’s path. Try to answer the questions below.

Questions:

For a fixed initial speed, at what launch angle is the maximum range achieved?
Keeping the launch angle at 45°, how does changing the initial speed affect the range and maximum height?
Turn on «Air Resistance». How does this change the shape of the trajectory compared to the ideal path?

Answer

The maximum range is achieved at a launch angle of 45°.
Increasing the initial speed increases both the range and the maximum height.
With air resistance, the trajectory is no longer a perfect parabola. The maximum height is lower, and the range is shorter than the ideal path. The projectile also descends at a steeper angle than it ascends.

👥 Pair/Group Work

Challenge your classmates with this projectile motion game on LearningApps!

Projectile Motion Group Quiz

💻 Individual Work (Structured Questions)

Analysis: A cannonball is fired with an initial velocity of 120 m/s at an angle of 55° above the horizontal. The cannon is located on a hill 50 m above a level plain. Calculate the total time the cannonball is in the air and the horizontal distance it travels before landing on the plain.
Analysis: A rescue plane flies at a constant velocity of 50 m/s at an altitude of 200 m. It drops a care package to stranded hikers. How far horizontally before the target point must the pilot release the package?
Synthesis: Two projectiles are launched from the same point at the same time. Projectile A is launched with an initial speed V at an angle of 30°. Projectile B is launched with the same initial speed V but at an angle of 60°. Ignoring air resistance, which projectile will have the greater range? Justify your answer mathematically.
Synthesis: A firefighter is aiming a hose at a fire on the third floor of a building, which is 15 m above the ground. The water leaves the hose with a speed of 20 m/s at an angle of 50° to the horizontal. The firefighter is standing 25 m away from the building. Will the water reach the fire? Provide a calculation to support your answer.
Analysis/Synthesis: A projectile is fired from the ground and lands on a target that is located at a horizontal distance R and a vertical height H. If the projectile is fired with an initial velocity u at an angle θ, derive an equation that relates R, H, u, and θ. This is often called the trajectory equation.

Answer

First, resolve components: u_y = 120*sin(55°) = 98.3 m/s; u_x = 120*cos(55°) = 68.8 m/s. Use y = u_yt + 0.5gt² with y = -50m. -50 = 98.3t — 4.905t². This is a quadratic equation: 4.905t² — 98.3t — 50 = 0. Solving for t gives t ≈ 20.5 s (the positive root). Range x = u_xt = 68.8 * 20.5 ≈ 1410 m.
First, find the time to fall 200m: y = 0.5gt² => 200 = 0.5 * 9.81 * t² => t ≈ 6.39 s. Then, find the horizontal distance: x = vt = 50 * 6.39 ≈ 319.5 m. The pilot must release the package about 320 m before the target.
The range is the same for both. The range formula is R = (u²sin(2θ))/g. For θ=30°, R = (u²sin(60°))/g. For θ=60°, R = (u²sin(120°))/g. Since sin(60°) = sin(120°), the ranges are identical. In general, for a given initial speed, the range is the same for launch angles θ and (90° — θ).
First, find the time to travel 25 m horizontally: t = x/u_x = 25 / (20*cos(50°)) ≈ 1.94 s. Now find the vertical height at that time: y = u_yt + 0.5gt² = (20*sin(50°))*1.94 + 0.5*(-9.81)*(1.94)² ≈ 29.7 — 18.4 ≈ 11.3 m. Since 11.3 m is less than the required 15 m, the water will not reach the fire.
Start with the two independent motion equations: x = (u*cosθ)t and y = (u*sinθ)t — 0.5gt². From the first equation, isolate t: t = x / (u*cosθ). Substitute this expression for t into the second equation: y = (u*sinθ)[x / (u*cosθ)] — 0.5g[x / (u*cosθ)]². Simplify to get the trajectory equation: y = x*tanθ — (gx²) / (2u²cos²θ). This equation relates the vertical height (y, or H) to the horizontal distance (x, or R).

🔗 Useful Links

🤔 Reflection

What is the most important concept to remember when solving any projectile motion problem? What part of today’s lesson was most challenging? Are there any scenarios of projectile motion you are still curious about?

Translation (Kazakh)

Кез келген снаряд қозғалысы есебін шешуде есте сақтау керек ең маңызды ұғым қандай? Бүгінгі сабақтың қай бөлігі ең қиын болды? Сізді әлі де қызықтыратын снаряд қозғалысының қандай да бір сценарийлері бар ма?