Motion in a Plane — Class 11 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Physics chapter "Motion in a Plane" — 7 important questions with detailed answers for CBSE board exam preparation.

TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Motion in a Plane" — 7 important questions with detailed answers for CBSE board exam p…

By Syllab.in · Updated Jun 14, 2026

Key Questions Covered:

Define projectile motion. A ball is thrown horizontally from a cliff of heigh…
A stone is projected at an angle of 30° with the horizontal with an initial v…
Define uniform circular motion. A particle moves in a circle of radius 5 m wi…
Two projectiles are fired from the same point with the same initial speed u b…
A car is moving in a circular path of radius 100 m with a constant speed of 2…
Define relative velocity. A boat moving at 10 m/s in still water wants to cro…
+ 1 more questions in the full chapter

Solutions Summary:

Question	Status
Define projectile motion. A ball is thrown horizontally f…	✓ Solved
A stone is projected at an angle of 30° with the horizont…	✓ Solved
Define uniform circular motion. A particle moves in a cir…	✓ Solved
Two projectiles are fired from the same point with the sa…	✓ Solved
A car is moving in a circular path of radius 100 m with a…	✓ Solved
Define relative velocity. A boat moving at 10 m/s in stil…	✓ Solved

Showing 6 of 7 questions

Q1: Define projectile motion. A ball is thrown horizontally from a cliff of height 45 m with initial velocity 20 m/s. Taking g = 10 m/s², find (a) time to reach ground (b) horizontal distance travelled (c) velocity when it hits the ground.

Projectile Motion Definition: Projectile motion is the motion of an object that is launched into the air and moves under the influence of gravity alone (neglecting air resistance). The path followed by a projectile is called a trajectory, which is parabolic in shape. Given: Height h = 45 m Initial horizontal velocity u_x = 20 m/s Initial vertical velocity u_y = 0 m/s (thrown horizontally) g = 10 m/s² Part (a): Time to reach ground Vertical motion: h = u_y × t + (1/2)gt² 45 = 0 + (1/2)(10)t² 4...

Q2: A stone is projected at an angle of 30° with the horizontal with an initial velocity of 40 m/s. Taking g = 10 m/s², find (a) maximum height (b) time of flight (c) range.

Given: Angle of projection θ = 30° Initial velocity u = 40 m/s g = 10 m/s² Resolve initial velocity: u_x = u cos θ = 40 cos 30° = 40 × (√3/2) = 20√3 m/s u_y = u sin θ = 40 sin 30° = 40 × (1/2) = 20 m/s Part (a): Maximum height At maximum height, v_y = 0 Using v_y² = u_y² - 2gH 0 = (20)² - 2(10)H 20H = 400 H = 20 m Part (b): Time of flight Time of flight T = 2u_y/g T = 2(20)/10 = 40/10 = 4 s Alternative: Using vertical motion y = u_y t - (1/2)gt² When projectile returns to ground level, y =...

Q3: Define uniform circular motion. A particle moves in a circle of radius 5 m with constant speed 10 m/s. Calculate (a) angular velocity (b) centripetal acceleration (c) period and frequency.

Uniform Circular Motion: Uniform circular motion is the motion of a body along a circular path at constant speed. Although the speed is constant, the velocity changes continuously in direction, so the body is accelerating toward the centre. This acceleration is called centripetal acceleration. Given: Radius r = 5 m Linear speed v = 10 m/s Part (a): Angular velocity Angular velocity ω = v/r ω = 10/5 = 2 rad/s Part (b): Centripetal acceleration Centripetal acceleration a_c = v²/r a_c = (10)²/...

Q4: Two projectiles are fired from the same point with the same initial speed u but at angles of 30° and 60° to the horizontal. Compare their ranges and maximum heights. Take g = 10 m/s².

Given: Both projectiles: initial speed = u, g = 10 m/s² Projectile 1: angle θ₁ = 30° Projectile 2: angle θ₂ = 60° Key observation: θ₂ = 90° - θ₁, so these are complementary angles. Part 1: Range comparison Range formula: R = u² sin 2θ/g For projectile 1 (30°): R₁ = u² sin(2 × 30°)/g = u² sin 60°/g = u² × (√3/2)/g For projectile 2 (60°): R₂ = u² sin(2 × 60°)/g = u² sin 120°/g = u² sin(180° - 60°)/g = u² sin 60°/g = u² × (√3/2)/g Therefore: R₁ = R₂ = u²√3/(2g) Conclusion: Both projectiles h...

Q5: A car is moving in a circular path of radius 100 m with a constant speed of 20 m/s. What is the centripetal force required if the mass of the car is 1000 kg? What provides this force?

Given: Radius of circular path r = 100 m Speed v = 20 m/s Mass of car m = 1000 kg Part 1: Calculate centripetal force Centripetal force F_c = mv²/r F_c = 1000 × (20)²/100 F_c = 1000 × 400/100 F_c = 1000 × 4 F_c = 4000 N = 4 kN Alternative calculation using angular velocity: ω = v/r = 20/100 = 0.2 rad/s F_c = mω²r = 1000 × (0.2)² × 100 = 1000 × 0.04 × 100 = 4000 N Part 2: What provides this force? The centripetal force is provided by the frictional force between the tyres and the road surfa...

Q6: Define relative velocity. A boat moving at 10 m/s in still water wants to cross a river flowing at 5 m/s. The river is 400 m wide. If the boat is directed perpendicular to the river current, find (a) the velocity of boat relative to ground (b) time to cross (c) how far downstream the boat will be when it reaches the opposite bank.

Relative Velocity Definition: Relative velocity is the velocity of an object as observed from a reference frame that is itself moving. It is the vector difference between the velocities of two objects. Given: Boat velocity in still water v_b = 10 m/s River current velocity v_r = 5 m/s River width = 400 m Boat directed perpendicular to river Part (a): Velocity of boat relative to ground When boat is directed perpendicular to the current: - Velocity component perpendicular to river (across) = v...

Showing 6 of 7 questions. Visit the full page for complete solutions.