NEET Questions on Projectile Motion
NEET Questions on Projectile Motion are from Chapter 4 Class 11 Physics Syllabus. Candidates are advised to solve the NEET MCQs on Projectile Motion regularly to ace the chapter.
NEET Questions on Projectile Motion involve questions based on Initial velocity, Trajectory, Horizontal velocity, Maximum height, Acceleration, Displacement, and Range. The NEET MCQ Questions on Projectile Motion will help students to know the question distribution asked in theNEET 2025 Examfrom this particular chapter. The weightage of the Important NEET UG Questions on Projectile Motion is 2%. It is expected that a total of 3 to 4 questions will be asked from the Projectile Motion in the NEET 2025 Syllabus. As suggested by toppers and experts, candidates must make sure to regularly attend theNEET Mock Test 2025to pass the exam without hurdles.
Important Questions on Projectile Motion will help students get an idea about the difficulty level of the MCQs asked from this section. The majority of the important NEET UG MCQ questions on Projectile Motion will include illustrations and diagrams. The questions are mainly derivative based. Students aiming to score good marks, must refer to theBest Books for NEET 2025 Preparation. The NEET Questions on Projectile Motion are included in the physics portion of the NEET Syllabus. Read the article below to learn in detail about the NEET Questions on Projectile Motion, important questions, and how to prepare.
Also Read:Should I Take a Drop for NEET 2025?
Important NEET Questions on Projectile Motion
The NEET Questions on Projectile Motion with solutions are mathematical in nature. Candidates must make sure that they regularly solve the previousNEET question paper pdfto strengthen their preparation. Here we have curated the important NEET questions on Projectile Motion with solutions for the reference of aspirants.
Q.1 Calculate the velocity with which an object is shot at an angle of 60° from the ground, and it reaches its maximum height in the 20s. Take g = 10 m/s2.
- 230.94 m/s
- 118 m/s
- 126 m/s
- 10.55 m/s
Answer: A
Explanation:
t = 40, θ = 60°, g = 10
time of flight will be in the 40s (time required for achieving maximum height is equal to half of the flight time)
Since, flight time is t = 2(v sinθ/g)
initial velocity will be 230.94 m/s
Q.2 An object of mass 2000 g covers a maximum vertical distance of 6 m when it is projected at an angle of 45° from the ground. Calculate the velocity with which it was thrown. Take g = 10 m/s2
- 12.10 m/s
- 15.49 m/s
- 2.155 m/s
- 12.0 m/s
Answer: A
Explanation:
h = 6, θ = 45°,
g = 10 and h = (v sinθ) 2/2g is the formula for maximum height.
By substituting the values, we get the initial velocity as 15.49 m/s
Q.3 What is the trajectory?
Answer:A trajectory is a curved path that is followed by a moving object through space as a function of time. A projectile that is thrown near the surface of the earth, the trajectory of the object is parabolic.
Q.4 What is the concept of acceleration in horizontal and vertical projectile motion?
Answer:The force that is acting on any object which is projected/thrown in the air with the some speed is the acceleration due to gravity. As the force acts vertically downwards, there is no acceleration in the horizontal direction. This means that the particle’s velocity remains constant in the horizontal direction.
Q. 5 In a normal projectile motion, what will be the condition for maximum range
- θ = 45°
- θ = 30°
- θ = 60°
- θ = 0°
Answer: A
Explanation:The formula for the horizontal range is R = v2(sin 2θ)/g
Hence, at sin 2θ = 1, the value of R will be maximum, therefore 2θ = 90°,
which means that θ should be 45°.
Hence the correct answer is θ = 45°
Q.6 In a normal projectile motion, what will be the condition for maximum height?
- θ = 55°
- θ = 90°
- θ = 180°
- θ = 0°
Answer: B
Explanation:The formula for height is h = (v sinθ)2/2g
Hence, at sin θ = 1, this will be the maximum,
θ should be 90°.
Hence the correct answer will be θ = 90°.
Q.7 In a normal projectile motion, which of the following quantities has an effect in calculating the body’s mass?
- Force
- Time
- Velocity
- Horizontal range
Answer: A
Explanation:The qualities that are mentioned above, are kinematic quantities. Force is the only kinetic energy.
Force = m x a is a kinetic quantity, the force is affected by the mass of the body.
Q.8 In a normal projectile motion, what will be the angle of the projectile to get the horizontal range minimum?
- θ = 90°
- θ = 30°
- θ = 45°
- θ = 75°
Answer: A
Explanation:R = v2 (sin 2θ)/g is the formula for the horizontal range
θ is taken as 90°
sin 2θ will be sin 180° = 0
Hence, the range covered will be 0
Q.9 An object of mass 2000 g covers a maximum vertical distance of 6 m when it is projected at an angle of 45° from the ground. Calculate the velocity with which it was thrown. Take g = 10 m/s2
- 12.10 m/s
- 15.49 m/s
- 2.155 m/s
- 12.0 m/s
Answer: A
Explanation:
h = 6, θ = 45°, g = 10, and h = (v sinθ) 2/2g
Using the values, we get the initial velocity as 15.49 m/s
Students who wish to attempt all the important NEET MCQ Questions on Projectile Motion are advised to regularly solve the previous year question papers. It is likely that a total of 2 to 3 questions will be asked in the NEET exam from this chapter. For more information related to medical, nursing, and paramedical, follow CollegeDekho.
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