Q.14 Physics ⭐ 360° Option Elimination Active

NEET 2023 Physics Question 14 Solution

Topic: Motion in a Plane · Official NTA Booklet Code G5

Question

A bullet is fired from a gun at the speed of 280 m s⁻¹ in the direction 30° above the horizontal. The maximum height attained by the bullet is (g = 9.8 m s⁻², sin 30° = 0.5):

✓

Verified Against NTA Final Answer Key

Option (4) : 1000 m

📚 Motion in a Plane Easy

📋 Given Parameters & Problem Setup

Initial Velocity (u) 280 m/s

Angle of Projection (θ) 30° above horizontal

Acceleration due to gravity (g) 9.8 m/s²

sin 30° 0.5

⚡

The 30-Second Exam Hall Shortcut: ⚡ Vertical Velocity Component Mental Shortcut

⏱️ Save ~40 seconds

u_y = u sin 30° = 280 × 0.5 = 140 m/s. Max height H = u_y² / (2g) = 140² / 19.6 = 19600 / 19.6 = 1000 m in 10 seconds!

1 Calculate the initial vertical velocity: u_y = u sin θ = 280 × 0.5 = 140 m/s.

2 Use standard kinematics at the apex (v_y = 0): H = u_y² / (2g).

3 Square 140: 140² = 19600.

4 Notice that 2g = 2 × 9.8 = 19.6.

5 Divide: 19600 / 19.6 = 1000 m exactly! Pick Option (4)!

🛡️

Why You Can 100% Trust This Shortcut: Fundamental Kinematics · Projectile Motion (Maximum vertical displacement under constant gravity: H = u² sin²θ / 2g).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Vertical Velocity Component Derivation

Resolve the initial velocity vector u along the vertical y-axis:

u_y = u \sin\theta = 280 \times \sin 30^\circ = 280 \times 0.5 = 140\text{ m/s}

Step 2

Maximum Height Formula Application

At the maximum height H, the vertical component of velocity momentarily drops to zero (v_y = 0). Applying the third equation of kinematics for vertical motion:

v_y^2 = u_y^2 - 2g H \implies 0 = u_y^2 - 2g H \implies H = \frac{u_y^2}{2g} = \frac{u^2 \sin^2\theta}{2g}

Step 3

Numerical Evaluation

Substitute the given values into the formula with g = 9.8 m/s²:

H = \frac{(140)^2}{2 \times 9.8} = \frac{19600}{19.6} = 1000\text{ m}

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 (1)

✗ Incorrect

Verdict & Scientific Reason:

Overestimates maximum height by a factor of 3.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate miscalculated powers of ten or confused with horizontal range.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Numerical distractor from multiplying speed by time.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate calculated u × t instead of u_y² / (2g).

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

Off by a factor of 2.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate forgot the factor of 2 in the denominator (calculated u_y² / g instead of u_y² / 2g).

4 (4)

✓ 100% Correct

Verdict & Scientific Reason:

Accurate value derived from (140)² / (2 × 9.8) = 19600 / 19.6 = 1000 m.

💡

Why this is airtight: None. Candidate executed the vertical component calculation flawlessly.

⚠️

NTA Trap Alert: Forgetting the Factor of 2 in 2g (Yielding 2000 m)

A frequent blunder is dividing by g (9.8) instead of 2g (19.6). This yields 19600 / 9.8 = 2000 m, matching trap Option (3).

🛡️ How to avoid the -1 mark: Check the denominator: Maximum height is u² sin²θ / (2g), whereas time of flight is 2u sinθ / g.

📖

Core NEET Syllabus Standard

National Eligibility cum Entrance Test (NEET UG) · Motion in a Plane Standard Benchmark

Core Scientific Principle Tested:

"Vertical motion of a projectile is independent of horizontal motion and governed strictly by u_y and g (H = u_y² / 2g)."

Syllabus Focus: Projectile Motion: Maximum Height, Range, and Flight Time

Preserved in Rationalized Curriculum Syllabus & Academic Standards ↗

Ready to attempt under real exam pressure?

Test your speed and accuracy in the authentic 200-minute timed NTA simulator.

Launch NEET 2023 CBT Test →

← Previous (Q.13) All Physics Questions Grid Next (Q.15) →

🔗 Related Questions — Motion in a Plane

All 5 Motion in a Plane PYQs →

NEET 2026 Q23

Total Linear Momentum Variation for Two Counter-Rotating Particles on a Circular Wire

View Solution →

NEET 2024 Q20

Vectors in Uniform Circular Motion

View Solution →

NEET 2021 Q41

Projectile Motion from an Accelerating Frame: Velocity and Acceleration upon Release

View Solution →

NEET 2021 Q43

Relating Uniform Circular Motion Speed to Projectile Maximum Height

View Solution →

Spotted a mistake in this solution? Report an error (support@mycompetitiveexams.com)

⚖️ Legal & Fair Dealing Notice: Sourced from publicly released NEET 2023 question papers and NEET answer keys for private study, research, and non-commercial educational practice under Fair Dealing (Section 52(1)(a) of the Indian Copyright Act, 1957). Not affiliated with or endorsed by NTA or NMC.

NEET 2023 Physics Question 14 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Vertical Velocity Component Mental Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Vertical Velocity Component Derivation

Maximum Height Formula Application

Numerical Evaluation

360° Option Elimination Engine Key Feature

NTA Trap Alert: Forgetting the Factor of 2 in 2g (Yielding 2000 m)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Motion in a Plane

Total Linear Momentum Variation for Two Counter-Rotating Particles on a Circular Wire

Vectors in Uniform Circular Motion

Projectile Motion from an Accelerating Frame: Velocity and Acceleration upon Release

Relating Uniform Circular Motion Speed to Projectile Maximum Height

Solve 4 NEET Questions a Day — One per Subject