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The standard projectile formulas, R = u² sin 2θ / g and T = 2u sin θ / g, only work when the object lands at the height it was launched from. Throw a stone straight out from a cliff and they give nonsense. The horizontal launch is simpler than the general case, once you treat the two directions separately. This post works it out with numbers and lists the checks worth running.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "Motion in a plane, projectile motion" in kinematics. The NEET (UG) 2026 syllabus lists projectile motion, and the JEE (Advanced) 2026 brochure lists "projectiles". The text is section 3.9 of the NCERT Class 11 Physics chapter, Motion in a Plane, whose Example 3.7 is a stone thrown horizontally from a cliff. Check the documents for your year.
Why can you split the motion in two?
NCERT treats the x and y components of projectile motion independently. With air resistance ignored, nothing pushes sideways, so horizontal velocity stays u for the whole flight. Vertically, the object starts with zero velocity and accelerates at g, exactly like a dropped stone. Put the origin at the edge of the cliff and take downward as positive for y:
Horizontal: x = ut
Vertical: y = ½gt², and vertical velocity v_y = gt
Every result below comes from these three lines.
Worked example 1: from a 45 m cliff
A stone is thrown horizontally at 12 m/s from the top of a 45 m cliff. Take g = 10 m/s².
Time of flight. It lands when y = 45 m: 45 = ½ × 10 × t², so t² = 9 and t = 3.0 s.
Horizontal distance. x = 12 × 3.0 = 36 m from the foot of the cliff.
Velocity at impact. Horizontal: 12 m/s. Vertical: v_y = 10 × 3.0 = 30 m/s. Speed = √(12² + 30²) = √1044 ≈ 32.3 m/s.
Direction. tan α = v_y / u = 30 / 12 = 2.5, so α ≈ 68.2° below the horizontal.
Energy check. v² = u² + 2gh = 144 + 900 = 1044. Same answer, with no time needed.
Path. Eliminating t gives y = gx² / (2u²) = 10x² / 288. At x = 36 m, y = 12960 / 288 = 45 m. It lands where it should.
Worked example 2: hitting a target
A rescue team on an 80 m cliff wants a supply pack to land on a boat 100 m from the foot of the cliff. At what horizontal speed should it leave the edge? Take g = 10 m/s².
Time is fixed by the height: t = √(2 × 80 / 10) = √16 = 4.0 s.
The pack must cover 100 m in 4.0 s: u = 100 / 4.0 = 25 m/s.
Impact: v_y = 40 m/s, speed = √(25² + 40²) = √2225 ≈ 47.2 m/s.
The method is always the same: height gives time, time links distance and speed.
Two quick checks worth knowing
Dropped versus thrown. Drop one stone and throw another horizontally from the same height at the same moment. Both land together, because their vertical motions are identical.
Angle shortcut. At impact, tan α = gt / u. Since h = ½gt² and R = ut, this equals 2h / R. In Example 1, 2 × 45 / 36 = 2.5, matching the direct result. Use it to check an answer in seconds.
Which mistakes cost marks?
Using R = u² sin 2θ / g. With θ = 0 it gives zero range. It does not apply here.
Giving the stone a starting vertical velocity. A horizontal throw has v_y = 0 at the start.
Adding speeds. Impact speed is √(u² + v_y²), not u + v_y (which would give 42 m/s in Example 1).
Measuring distance from the wrong point. x is measured from the foot of the cliff.
Mixing sign conventions. Pick downward or upward as positive and keep it for every equation.
A practice routine
Rework NCERT Example 3.7 (490 m, 15 m/s, g = 9.8 m/s²) without looking. You should get 10 s and about 99 m/s.
Redo Example 1 with u = 24 m/s. Predict before calculating: same time, double the distance.
Solve one problem where the range is given and the height is asked.
Move on to angled launches with our posts on range and maximum height and projectiles on an incline.
Seeing the two motions side by side helps. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read, so you can change the launch speed and watch the fall time stay the same.
Before your next practice set, solve Example 2 again for a boat 140 m away.
