Projectile Motion
Study the motion of projectiles, including the independence of horizontal and vertical components, trajectory equations, maximum height, and the effect of initial conditions.
Why this shows up in the exam
Projectile motion is a classic NEET problem area, requiring you to analyze two-dimensional motion using kinematic principles.
How NEET tests this
Learn the idea
Projectile motion is two‑dimensional motion where the horizontal and vertical components act independently under constant horizontal velocity and constant vertical acceleration g. The key insight is that any required angle or distance can be obtained by treating the two components separately and then recombining them with simple geometry.
🧠 Memory hook: Think of a projectile as a car moving straight ahead (horizontal) while a ball drops from its roof (vertical); the car’s speed never changes, the ball’s speed changes uniformly – combine the two to get the path.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- Horizontal displacement: x = u cosθ · t (constant horizontal velocity)
- Vertical displacement: y = u sinθ · t – ½ g t² (uniformly accelerated)
- Maximum height: H = u² sin²θ / (2g)
- Range: R = u² sin2θ / g
- Time of flight: T = 2u sinθ / g
- Trajectory equation: y = x tanθ – (g x²)/(2u² cos²θ)
How to approach it
- 1Read the question and note the given initial speed u, launch angle θ and g.
- 2Write the required quantity in terms of the standard projectile formulas (H, R, T, etc.) using the independent horizontal and vertical components.
- 3If the problem involves an angle measured from the launch point, form a right‑triangle with base = horizontal distance and height = vertical distance, then use tan α = opposite/adjacent.
- 4Plug the numbers, simplify and choose the matching option.
Worked example — watch it click
A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is:
- A)60°
- ✅tan⁻¹ 1/2
- C)tan⁻¹ √3/2
- D)45°
The concept behind this problem
The worked example asks for the elevation angle to the highest point, so you must relate the apex height H to the half‑range R/2; this forces you to use both the height and range formulas, testing the independence of components and geometric recombination.
Step by step
- 1Projectile fired at 45°.
- 2At highest point, horizontal distance R/2 = (u²sin90°)/(2g) = u²/(2g).
- 3Maximum height H = u²sin²45°/(2g) = u²/(4g).
- 4Elevation angle from projection point: tan α = H/(R/2) = [u²/(4g)]/[u²/(2g)] = (1/4)/(1/2) = 1/2.
- 5Therefore α = tan⁻¹(1/2).
Watch out
Students often take tan α = H / R instead of the correct tan α = H / (R/2), giving the wrong angle.
Common slip-ups that cost marks
- •Confusing the angle of the velocity vector at the top (zero) with the elevation angle from the launch point to the apex.
- •Using sinθ where cosθ is needed (or vice‑versa) in the range or height formulas.
- •Forgetting that the horizontal component never changes, so the time used must be the same for both components.
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Practise it
These are real questions from past NEET papers that test this exact idea.
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)
Push further
More challenging3 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.
A cricket ball is hit with an initial speed of 35 m/s at an angle of 45° with the horizontal. If air resistance is negligible and g = 10 m/s², what is the maximum height achieved by the ball?
More from Kinematics
Distance, Displacement, Speed, and Velocity
Understand the differences between distance and displacement, and between speed and velocity, including how to calculate average speed and average velocity in various scenarios.
Equations of Motion and Uniform Acceleration
Learn the kinematic equations for uniformly accelerated motion, including applications to free fall, retardation, and calculation of distance in specific time intervals.
Relative Velocity and Motion Analysis
Explore how to determine the velocity of one object relative to another and analyze motion from different reference frames, including periodic motion.
Position, Path Length, and Displacement
For one-dimensional motion, displacement over an interval is Delta x = x_f - x_i and may be positive, negative, or zero. Distance is the non-negative path length, so distance is always at least |Delta x|.
Speed and Velocity
Instantaneous velocity is the signed rate v = dx/dt, while instantaneous speed is |v|. In one dimension the sign of v identifies motion along or opposite the positive axis.
Average Speed and Average Velocity
Over elapsed time Delta t, average speed equals total distance/Delta t and average velocity equals Delta x/Delta t. For equal distances the relevant mean of speeds is harmonic, not arithmetic.