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#Projectile Motion

158 public questions tagged with this topic.

PHYSICS

A projectile is launched at 20 m/s at 53° . What is its maximum height? (Take g = 10 m/s², sin 53° = 0.8 )

Given: A projectile is launched at 20 m/s at 53° . What is its maximum height? (Take g = 10 m/s², sin 53° = 0.8 ) These values define the system as per NCERT data. Formula: Maximum height h_m = (v_0 sin θ_0)²/2g. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: v_0 = 20 m/s, sin 53° = 0.8, g = 10 m/s² . h_m = (20 × 0.8)²/2 × 10 = 16²/20 = 256/20 = 12.8 m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NC

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A 1700 kg truck accelerates at 1.5 m/s² for 4 s, then a 1.8 kg stone is dropped. What is the horizontal distance travele

Given: A 1700 kg truck accelerates at 1.5 m/s² for 4 s, then a 1.8 kg stone is dropped. What is the horizontal distance traveled by the stone in 2.5 s ? (Neglect air resistance) Formula: Truck’s velocity after 4 s : v = u + at = 0 + 1.5 × 4 = 6 m/s. Substitution & Calculation: Stone inherits this horizontal velocity upon drop. No horizontal forces act after drop (air resistance neglected). Distance: s = v × t = 6 × 2.5 = 15 m . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A stone is dropped from a cliff 45 m high with an initial horizontal speed of 15 m/s. What is its speed on hitting the g

Given: A stone is dropped from a cliff 45 m high with an initial horizontal speed of 15 m/s. What is its speed on hitting the ground? (Take g = 10 m/s²) These values define the system as per NCERT data. Formula: Horizontal velocity: v_x = 15 m/s (constant). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Vertical velocity: v_y = √(2gh) = √(2 × 10 × 45) = √900 = 30 m/s. Speed v = √(v_x² + v_y²) = √(15² + 30²) = √(225 + 900) = √1125 ≈ 33.54 m/s. Result: The computed value matches the expected outcome and confirms the corr

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A stone is thrown horizontally at 15 m/s from a height of 19.6 m . How much time does it take to reach the ground? (Take

Given: A stone is thrown horizontally at 15 m/s from a height of 19.6 m . How much time does it take to reach the ground? (Take g = 9.8 m/s² ) These values define the system as per NCERT data. Formula: Vertical motion: y = 1/2 g t², where y = -19.6 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: -19.6 = 1/2 × 9.8 × t² Rightarrow -19.6 = 4.9 t² Rightarrow t² = 4 Rightarrow t = 2 s . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are prop

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

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PHYSICS

A ball is thrown upwards at 20 m/s from a 50 m tower. What is its speed when it passes 20 m below the tower’s top? (Take

Given: A ball is thrown upwards at 20 m/s from a 50 m tower. What is its speed when it passes 20 m below the tower’s top? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Displacement y = -20 m, v² = (20)² + 2 · 10 · 20 = 400 + 400 = 800. This is standard NCERT relation. Substitution & Calculation: v = sqrt800 approx 28.28 m/s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A ball is thrown horizontally at 6 m/s from a height of 4.9 m . What is its horizontal range? (Take g = 9.8 m/s² )

Given: A ball is thrown horizontally at 6 m/s from a height of 4.9 m . What is its horizontal range? (Take g = 9.8 m/s² ) Formula: Time to fall: h = 1/2 g t² Rightarrow 4.9 = 1/2 × 9.8 × t² Rightarrow 4.9 = 4.9 t² Rightarrow t² = 1 Rightarrow t = 1 s. Substitution & Calculation: Horizontal range R = v_x t = 6 × 1 = 6 m . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A projectile is launched at 45 m/s at 30° . What is its speed at maximum height? (Take g = 10 m/s², cos 30° = 0.866 )

Given: A projectile is launched at 45 m/s at 30° . What is its speed at maximum height? (Take g = 10 m/s², cos 30° = 0.866 ) These values define the system as per NCERT data. Formula: At maximum height, speed = v_x = v_0 cos θ_0. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: v_0 = 45 m/s, cos 30° = 0.866 . v_x = 45 × 0.866 approx 38.97 m/s . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

Two balls are thrown upwards simultaneously from the ground with speeds 20 m/s and 30 m/s . After how much time do they

Given: Two balls are thrown upwards simultaneously from the ground with speeds 20 m/s and 30 m/s . After how much time do they have the same velocity? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Velocity: v_1 = 20 - 10 t, v_2 = 30 - 10 t. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Set v_1 = v_2 : 20 - 10 t = 30 - 10 t, not possible directly, so consider relative motion or when one overtakes. Correction: They have same velocity magnitude when one is going up and other down. Time to max height: t_1 = 2 s, t_

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

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PHYSICS

A ball is thrown upwards at 25 m/s from a 60 m cliff. What is its speed when it hits the ground? (Take g = 10 m/s² )

Given: A ball is thrown upwards at 25 m/s from a 60 m cliff. What is its speed when it hits the ground? (Take g = 10 m/s² ) Formula: v² = v_0² + 2 g h, v² = (25)² + 2 · 10 · 60 = 625 + 1200 = 1825. Substitution & Calculation: v = √1825 approx 42.72 m/s . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A ball is thrown upwards with a speed of 38 m/s . What is its velocity after 3 s ? (Take g = 10 m/s² )

Given: A ball is thrown upwards with a speed of 38 m/s . What is its velocity after 3 s ? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Use v = v_0 + a t. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Here, v_0 = 38 m/s, a = -10 m/s², t = 3 s . Substitute: v = 38 - 10 · 3 = 38 - 30 = 8 m/s . The velocity is 8 m/s . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A ball is thrown upwards with a speed of 60 m/s . What is the time to reach the maximum height? (Take g = 10 m/s² )

Given: A ball is thrown upwards with a speed of 60 m/s . What is the time to reach the maximum height? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: At max height, v = 0. This is standard NCERT relation. Substitution & Calculation: Use v = v_0 + a t . Here, v_0 = 60 m/s, a = -10 m/s² . Substitute: 0 = 60 - 10 t Rightarrow 10 t = 60 Rightarrow t = 6 s . The time taken is 6 s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A ball is thrown vertically upwards with an initial speed of 28 m/s . What is the total time of flight? (Take g = 10 m/s

Given: A ball is thrown vertically upwards with an initial speed of 28 m/s . What is the total time of flight? (Take g = 10 m/s² ) Formula: Time to max height: v = v_0 + a t, 0 = 28 - 10 t Rightarrow t = 2.8 s. Substitution & Calculation: Total time = time up + time down = 2.8 s + 2.8 s = 5.6 s . The total time is 5.6 s . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27), Chapter: Motion in a Straight Line (Latest NCERT 2026-27), Topic: Kinematic equations, a = (v-v₀)/t, x = v₀t + ½at² and distance. The secti

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