This MCQ module is based on: Projectile Circular Motion
TOPIC 12 OF 33
Projectile Circular Motion
🎓 Class 11
Physics
CBSE
Theory
Ch 3 – Motion in a Plane
⏱ ~14 min
🌐 Language: [gtranslate]
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Projectile Circular Motion
3.10 Projectile Motion
As an application of the ideas developed in the previous sections, we consider the motion of a projectile. An object that is in flight after being thrown or projected is called a projectile. Such a projectile might be a football, a cricket ball, a baseball or any other object. The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motions:
- Horizontal motion: uniform velocity, no acceleration
- Vertical motion: constant acceleration \(\vec{g}\) due to gravity (downward)
It was Galileo who first stated this independence principle. In our discussion, we shall assume that the air resistance has negligible effect on the motion of the projectile.
Suppose that the projectile is launched with velocity \(\vec{v_0}\) that makes an angle \(\theta_0\) with the x-axis. After the object has been projected, the acceleration acting on it is that due to gravity:
\[\vec{a} = -g\,\hat{j}\]
The components of initial velocity \(\vec{v_0}\) are:
\[v_{0x} = v_0\cos\theta_0, \quad v_{0y} = v_0\sin\theta_0\]
Equations of Projectile Motion
The position of the projectile at time t (taking g = 9.8 m/s²):
\[x(t) = (v_0\cos\theta_0)\,t\]
\[y(t) = (v_0\sin\theta_0)\,t - \frac{1}{2}g\,t^2\]
Velocity components:
\[v_x = v_0\cos\theta_0 \text{ (constant)}\]
\[v_y = v_0\sin\theta_0 - g\,t\]
Equation of the Trajectory (Parabola)
Eliminating t from the position equations: \(t = x/(v_0\cos\theta_0)\). Substituting in y:
\[\boxed{y = (\tan\theta_0)\,x - \frac{g}{2(v_0\cos\theta_0)^2}\,x^2}\]
This is the equation of a parabola. So the trajectory of a projectile is a parabola.
Time of Flight (T)
The projectile returns to the same height (y = 0) at time t = T. Setting y = 0:
\[T = \frac{2v_0\sin\theta_0}{g}\]
Maximum Height (h_m)
At max height, \(v_y = 0\). So time to reach max height = T/2. Using \(v_y^2 = v_{0y}^2 - 2gh_m\):
\[h_m = \frac{(v_0\sin\theta_0)^2}{2g} = \frac{v_0^2\sin^2\theta_0}{2g}\]
Horizontal Range (R)
The horizontal distance covered when projectile returns to same height:
\[R = (v_0\cos\theta_0) \cdot T = \frac{v_0^2 \cdot 2\sin\theta_0\cos\theta_0}{g} = \frac{v_0^2\sin 2\theta_0}{g}\]
Maximum Range: Since \(\sin 2\theta_0\) is max when \(2\theta_0 = 90°\), the range is maximum when \(\theta_0 = \boxed{45°}\):
\[R_{max} = \frac{v_0^2}{g}\]
Worked Example 1 (NCERT Example 3.7): Projectile from Mountain
Galileo, in his book Two New Sciences, stated that "for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal." Prove this statement.
Proof: Range formula: \(R = \frac{v_0^2 \sin 2\theta_0}{g}\).
Take two angles: \(\theta_1 = 45° + \alpha\) and \(\theta_2 = 45° - \alpha\).
Then \(2\theta_1 = 90° + 2\alpha\) and \(2\theta_2 = 90° - 2\alpha\).
\[\sin(90° + 2\alpha) = \cos(2\alpha) = \sin(90° - 2\alpha)\] Therefore \(R_1 = R_2\). ∎
Example: 30° and 60° give the same range; 20° and 70° give the same range.
Take two angles: \(\theta_1 = 45° + \alpha\) and \(\theta_2 = 45° - \alpha\).
Then \(2\theta_1 = 90° + 2\alpha\) and \(2\theta_2 = 90° - 2\alpha\).
\[\sin(90° + 2\alpha) = \cos(2\alpha) = \sin(90° - 2\alpha)\] Therefore \(R_1 = R_2\). ∎
Example: 30° and 60° give the same range; 20° and 70° give the same range.
Worked Example 2 (NCERT Example 3.8): Cricket Ball Range
A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with initial speed of 15 m/s. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground (g = 9.8 m/s²).
Setup: Take cliff edge as origin, x horizontal (direction of throw), y downward as +.
\(v_{0x} = 15\) m/s, \(v_{0y} = 0\), drop height = 490 m, g = 9.8 m/s².
Time to reach ground: Vertical motion: \(490 = \frac{1}{2}(9.8)t^2\) \[t^2 = \frac{2 \times 490}{9.8} = 100 \Rightarrow t = \boxed{10 \text{ s}}\]
Speed when hitting ground: \[v_x = 15 \text{ m/s (unchanged)}\] \[v_y = g\,t = 9.8 \times 10 = 98 \text{ m/s}\] \[|\vec{v}| = \sqrt{15^2 + 98^2} = \sqrt{225 + 9604} = \sqrt{9829} \approx \boxed{99.1 \text{ m/s}}\]
\(v_{0x} = 15\) m/s, \(v_{0y} = 0\), drop height = 490 m, g = 9.8 m/s².
Time to reach ground: Vertical motion: \(490 = \frac{1}{2}(9.8)t^2\) \[t^2 = \frac{2 \times 490}{9.8} = 100 \Rightarrow t = \boxed{10 \text{ s}}\]
Speed when hitting ground: \[v_x = 15 \text{ m/s (unchanged)}\] \[v_y = g\,t = 9.8 \times 10 = 98 \text{ m/s}\] \[|\vec{v}| = \sqrt{15^2 + 98^2} = \sqrt{225 + 9604} = \sqrt{9829} \approx \boxed{99.1 \text{ m/s}}\]
Worked Example 3 (NCERT Example 3.9): Football Kick
A cricket ball is thrown at a speed of 28 m/s in a direction 30° above the horizontal. Calculate (a) maximum height, (b) time taken by the ball to return to the same level, (c) distance from the thrower to the point where the ball returns to the same level.
Given: v₀ = 28 m/s, θ₀ = 30°, g = 9.8 m/s².
(a) Max height: \[h_m = \frac{v_0^2 \sin^2\theta_0}{2g} = \frac{(28)^2 (0.5)^2}{2 \times 9.8} = \frac{784 \times 0.25}{19.6} = \boxed{10.0 \text{ m}}\]
(b) Time of flight: \[T = \frac{2v_0\sin\theta_0}{g} = \frac{2 \times 28 \times 0.5}{9.8} = \frac{28}{9.8} \approx \boxed{2.86 \text{ s}}\]
(c) Range: \[R = \frac{v_0^2 \sin 2\theta_0}{g} = \frac{(28)^2 \times \sin60°}{9.8} = \frac{784 \times 0.866}{9.8} \approx \boxed{69.3 \text{ m}}\]
(a) Max height: \[h_m = \frac{v_0^2 \sin^2\theta_0}{2g} = \frac{(28)^2 (0.5)^2}{2 \times 9.8} = \frac{784 \times 0.25}{19.6} = \boxed{10.0 \text{ m}}\]
(b) Time of flight: \[T = \frac{2v_0\sin\theta_0}{g} = \frac{2 \times 28 \times 0.5}{9.8} = \frac{28}{9.8} \approx \boxed{2.86 \text{ s}}\]
(c) Range: \[R = \frac{v_0^2 \sin 2\theta_0}{g} = \frac{(28)^2 \times \sin60°}{9.8} = \frac{784 \times 0.866}{9.8} \approx \boxed{69.3 \text{ m}}\]
🎯 Interactive Simulation: Projectile Launcher
Adjust launch speed and angle. Observe the parabolic trajectory and compare ranges.
Test Galileo's claim: Try θ = 30° and θ = 60° — same range!
3.11 Uniform Circular Motion
When an object follows a circular path at constant speed, the motion of the object is called uniform circular motion (UCM).
Centripetal Acceleration
Even though the speed is constant in UCM, the velocity vector is constantly changing direction. This change of velocity produces an acceleration directed towards the center of the circle, called centripetal acceleration.
Centripetal Acceleration Magnitude:
\[a_c = \frac{v^2}{r} = \omega^2 r = v\omega\]
where v = speed, r = radius, \(\omega\) = angular speed (rad/s).
Angular Speed and Period
If the object completes one full revolution in time T (the period):
\[v = \frac{2\pi r}{T}, \quad \omega = \frac{2\pi}{T} = 2\pi f\]
where f = 1/T is the frequency (revolutions per second, Hz).
Worked Example 4 (NCERT Example 3.10): Insect on Turntable
An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed and the linear speed of the motion? (b) Is the acceleration vector a constant vector? What is its magnitude?
Given: r = 12 cm = 0.12 m, 7 revolutions in 100 s.
(a) Time per revolution: T = 100/7 ≈ 14.3 s.
\[\omega = \frac{2\pi}{T} = \frac{2\pi \times 7}{100} = 0.44 \text{ rad/s}\] \[v = \omega r = 0.44 \times 0.12 = \boxed{0.053 \text{ m/s}}\]
(b) The acceleration vector is NOT constant — its magnitude is constant but direction continuously changes (always toward center). Its magnitude: \[a_c = \omega^2 r = (0.44)^2 \times 0.12 = \boxed{0.023 \text{ m/s}^2}\] or equivalently \(a_c = v^2/r = (0.053)^2 / 0.12 = 0.023\) m/s².
(a) Time per revolution: T = 100/7 ≈ 14.3 s.
\[\omega = \frac{2\pi}{T} = \frac{2\pi \times 7}{100} = 0.44 \text{ rad/s}\] \[v = \omega r = 0.44 \times 0.12 = \boxed{0.053 \text{ m/s}}\]
(b) The acceleration vector is NOT constant — its magnitude is constant but direction continuously changes (always toward center). Its magnitude: \[a_c = \omega^2 r = (0.44)^2 \times 0.12 = \boxed{0.023 \text{ m/s}^2}\] or equivalently \(a_c = v^2/r = (0.053)^2 / 0.12 = 0.023\) m/s².
📐 Activity 3.4 — Spinning Bucket Demonstration
L3 Apply
Demonstration setup: Tie a small bucket of water to a sturdy rope (length ~1 m). Spin the bucket in a vertical circle quickly enough so the water doesn't spill — even when the bucket is upside down at the top!
Predict: What is the minimum speed at the top so water doesn't fall out? Use the centripetal acceleration concept.
At the top of the circle, the bucket is upside down. The forces on the water are:
- Gravity (mg) — pointing down (toward center)
- Normal force from bucket bottom — pointing down (toward center)
For the water to stay in the bucket, gravity alone must provide AT LEAST the centripetal force:
\[mg \le \frac{mv^2}{r} \Rightarrow v \ge \sqrt{gr}\]
For r = 1 m: v_min = √(9.8 × 1) ≈ 3.13 m/s.
Conclusion: If the bucket moves slower than 3.13 m/s at the top, gravity exceeds the required centripetal force and the water falls. This is why you must spin fast enough!
🎯 Competency-Based Questions
A javelin is thrown by an athlete at an angle of 35° to the horizontal with initial speed 25 m/s. The track stadium is at sea level (g = 9.8 m/s²).
Q1. Calculate the maximum height attained by the javelin. L3 Apply
Answer: \(h_m = \frac{v_0^2\sin^2\theta_0}{2g} = \frac{625 \times \sin^2(35°)}{19.6} = \frac{625 \times 0.329}{19.6} \approx \boxed{10.49 \text{ m}}\).
Q2. Calculate the horizontal range of the javelin. L3 Apply
Answer: \(R = \frac{v_0^2 \sin 2\theta_0}{g} = \frac{625 \times \sin 70°}{9.8} = \frac{625 \times 0.94}{9.8} \approx \boxed{59.93 \text{ m}}\).
Q3. The athlete in Q1 wants to maximize range. By how much should they change the angle? Justify with computation. L4 Analyse
Answer: Maximum range occurs at θ = 45°. So the athlete should INCREASE the angle by 10°. New range: \(R_{max} = v_0^2/g = 625/9.8 \approx \boxed{63.78 \text{ m}}\). Improvement: 63.78 − 59.93 ≈ 3.85 m. (Note: in real javelin, optimum is ~30–35° due to aerodynamics, but for ideal projectile, 45° is best.)
Q4. A car moves around a curve of radius 50 m at constant speed 20 m/s. Calculate centripetal acceleration. L3 Apply
Answer: (b) 8 m/s². \(a_c = v^2/r = 400/50 = 8\) m/s², directed toward center of curve.
Q5. HOT (Create): Design a roller coaster loop. The loop has radius 15 m. What's the MINIMUM speed required at the top of the loop so passengers don't lose contact with the track? Justify by listing the forces and computing. L6 Create
Sample Solution:
At the top of the loop, the track is above the passenger. Forces on passenger:
- Gravity (mg) — DOWN (toward center of loop)
- Normal from track (N) — DOWN (toward center)
For circular motion: \(mg + N = \frac{mv^2}{r}\). Minimum speed when N = 0 (just losing contact):
\[v_{min} = \sqrt{gr} = \sqrt{9.8 \times 15} = \sqrt{147} \approx \boxed{12.12 \text{ m/s}}\]
So roller coaster must be designed for at least ~12 m/s at the top. (Typically engineers add safety factor ≥ 1.4 → minimum design speed ~17 m/s.)
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: The trajectory of a projectile is always a parabola.
R: Horizontal velocity is constant and vertical motion has constant downward acceleration.
Answer: (A). Both true; R explains A. Eliminating t from x(t) and y(t) yields y = (tan θ)x − [g/(2v₀²cos²θ)]x², a parabola.
A: Two projectiles thrown at angles 30° and 60° with the same speed have the same range.
R: Range R = (v₀²sin 2θ)/g, and sin(2×30°) = sin(2×60°) = sin 60°.
Answer: (A). Both true; R is the exact reason. \(\sin 60° = \sin 120°\), so ranges match (though heights and times differ).
A: A particle in uniform circular motion has constant velocity.
R: The speed of the particle remains constant throughout the motion.
Answer: (D). A is FALSE — velocity (a vector) keeps changing direction, hence not constant. R is TRUE — speed (magnitude only) IS constant. The distinction between speed (scalar) and velocity (vector) is essential.
Frequently Asked Questions - Projectile Circular Motion
What is the main concept covered in Projectile Circular Motion?
In NCERT Class 11 Physics Chapter 3 (Motion in a Plane), "Projectile Circular Motion" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Projectile Circular Motion useful in real-life applications?
Real-life applications of Projectile Circular Motion from NCERT Class 11 Physics Chapter 3 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Projectile Circular Motion?
Key formulas in Projectile Circular Motion (NCERT Class 11 Physics Chapter 3 Motion in a Plane) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 3?
NCERT Class 11 Physics Chapter 3 (Motion in a Plane) is structured so each part builds on the previous one. Projectile Circular Motion connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Projectile Circular Motion?
CBSE board questions from Projectile Circular Motion typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Projectile Circular Motion lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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