Projectile Motion: Check the Landing Level First

Projectile motion is the path of an object thrown into the air and then moving under gravity alone. It splits into two independent motions: a horizontal one with zero acceleration and a vertical one with acceleration g = 9.8 m/s² downward. The standard treatment ignores air resistance (NCERT Class 11 Physics, Chapter 3).

Let u be the launch speed and θ the angle above the horizontal.

• Launch components: ux = u cos θ and uy = u sin θ.
• Position at time t: x = u cos θ × t and y = u sin θ × t − ½gt².
• Velocity at time t: vx = u cos θ and vy = u sin θ − gt.

Putting y = 0 gives three results. NCERT derives them this way, and we call them the shortcuts:

• Time of flight, T = 2u sin θ / g.
• Maximum height, H = u² sin²θ / 2g.
• Horizontal range, R = u² sin 2θ / g, with a maximum of u²/g at 45°.

They are quick because they were derived for one special case: a throw that lands at the height it started from. Outside that case they do not apply. OpenStax states it directly: the range formula does not apply when the initial and final y positions differ.

Use the shortcuts only when the object lands at the height it was launched from. For anything else, such as a stone thrown off a cliff, write the position equations and solve for the time first. Then find the horizontal distance and the final speed from that time.

Run these three checks on every question.

Does the object land at the height it started from?

Subtract the starting height from the landing height. A result of zero passes, and you may use the shortcuts. Any other value fails: use the position equations. A cliff, a roof or a pit all fail.

Is the launch split into components correctly?

A pass means that for an angle measured from the horizontal, ux = u cos θ and uy = u sin θ. A horizontal throw (θ = 0) then gives uy = 0 and ux = u. If the question gives the angle with the vertical, swap them: ux = u sin θ. A fail is a horizontal throw that starts with a vertical speed.

Does the answer survive the sanity test?

For a level-ground throw, pass if all four hold:

• R is no more than u²/g.
• R equals ux × T.
• R equals 4H / tan θ (this is 4H at 45°).
• R is in metres, not seconds.

Any break means you have a slip. Our 3-step units check catches the last one.

On level ground, R depends on sin 2θ, which peaks at 2θ = 90°, so 45° gives the longest range. Angles that sit equally far either side of 45° give equal ranges. NCERT Example 3.6 proves it: for 45° + α and 45° − α, sin(90° + 2α) = sin(90° − 2α). The worked examples on this page are NCERT's. If you use the Maharashtra State Board (Balbharati) Std XI Physics textbook, its Motion in a Plane chapter covers the same projectile results.

Here is a table for a launch speed of 28 m/s, the speed in NCERT Example 3.8, with g = 9.8 m/s². We computed every cell from the three shortcuts, so a calculator reproduces it.

Read three things from it:

• 15° and 75°, 30° and 60°, and 40° and 50° each tie on range.
• Complementary angles share a range but not a path. At 60° the height is three times the 30° height (30.0 m against 10.0 m) and the flight is √3 times longer (4.95 s against 2.86 s). OpenStax makes the same point: complementary angles give equal range and different heights.
• At 45°, R = 80.0 m = 4 × 20.0 m, and u²/g = 28² / 9.8 = 80.0 m. This is the sanity test in action.

At the highest point the vertical velocity is zero, the horizontal velocity is unchanged and the acceleration is g downward. The vertical velocity is only passing through zero on its way from upward to downward. A velocity that is zero for an instant does not mean the acceleration is zero.

Take the 28 m/s throw at 30°:

• The vertical launch component is 28 sin 30° = 14 m/s. It reaches zero after 14 / 9.8 ≈ 1.43 s, which is half of T = 2.86 s.
• The horizontal velocity stays at 28 cos 30° ≈ 24.2 m/s all the way, because the horizontal acceleration is zero (NCERT Chapter 3, ax = 0 and ay = −g).
• So the speed at the top is about 24.2 m/s, not zero.

When you throw from a height, 45° is no longer the best angle, because the extra fall time adds horizontal distance whatever the angle. Favouring horizontal speed then pays off, so the best angle drops below 45°. This is our own working, and it holds in the no-air-resistance model.

For a launch speed u from height h, the range-maximising angle satisfies tan θ = u / √(u² + 2gh). We checked this by computing the range for every whole-degree angle from 1° to 89°. The table uses u = 20 m/s and g = 9.8 m/s².

The best angle falls steadily, but 45° is still close in range: it loses 0.7 m, 1.9 m and 5.8 m at the three heights. Exam options usually carry that wrinkle, so a question that says "from a height" should not get the answer 45°.

NCERT Example 3.7 is the cleanest cliff case. A stone is thrown horizontally at 15 m/s from a cliff 490 m high.

1. The stone lands when −490 = −½ × 9.8 × t², so t = 10 s.
2. Its vertical velocity at landing is 9.8 × 10 = 98 m/s downward.
3. Its speed is √(15² + 98²) ≈ 99 m/s, as NCERT gives.
4. It travels 15 × 10 = 150 m horizontally (our arithmetic).

The shortcut would have given R = u² sin 0° / g = 0, which is plainly wrong. This is why the first of the three checks is whether the object lands at its launch height.

Compare your answer with u²/g, with ux × T, and with 4H / tan θ. A level-ground range can never exceed u²/g. Here is an example with a launch of 40 m/s at 50° and g = 9.8 m/s², worked with the shortcuts.

• T = 2 × 40 × sin 50° / 9.8 ≈ 6.25 s
• H = (40 sin 50°)² / (2 × 9.8) ≈ 47.9 m
• R = 40² × sin 100° / 9.8 ≈ 160.8 m
• u²/g = 1600 / 9.8 ≈ 163.3 m, so R ≤ u²/g holds
• ux × T = 40 cos 50° × 6.25 ≈ 25.7 × 6.25 ≈ 160.8 m, which matches

A range of 6.25 m fails at once. It would need a horizontal speed of about 1 m/s, but ux here is 25.7 m/s. A slip to watch for is giving T's number as R, and the "metres, not seconds" check catches it.

For a step-by-step method on any numerical, read our 4-step routine for HSC and MHT-CET.

Projectile motion belongs to Motion in a plane, one of the ten Std XI Physics chapters in the MHT-CET syllabus. The CET Cell's MHT-CET 2026 syllabus, listed for A.Y. 2026-2027 and published on 21 November 2025, gives Std XI about 20% of the paper. Its Physics table lists 10 Std XI and 40 Std XII questions at 1 mark each, with no negative marking.

The same document says questions are "mainly application based" and set at JEE (Main) difficulty for Physics. Our recommendation follows from that: expect the landing-level condition to be hidden in the wording, such as a launch from a tower or a landing in a pit. For why question counts per chapter are estimates and not official, read our note on why chapter weightage is historical, not official.

Our MHTCET 2028 – Class 11th Physics Batch, taught by Rahul Jewaani (RG Sir), covers the complete Class 11 syllabus in depth, with a bridge from Class 10 to Class 11. Run the three checks above on every projectile question you practise.

As of October 2026, the batch costs ₹1,212 and is valid until 30/06/2028. If you are in Class 12 and skipped this chapter, start with our catch-up plan for Class 11 MHT-CET Physics.