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A projectile's path is a parabola, yet exam questions sometimes ask for "the radius of curvature at the highest point" as if it were a circle. The idea is that every short piece of a smooth curve matches a circle, and the circle's radius tells you how sharply the path bends there. You can find it with nothing more than the centripetal acceleration formula. This post shows where the formula comes from, works one throw at three points along its path, and lists the slips to avoid.
Where does this sit in the syllabus?
The current NCERT Class 11 Physics Part I chapter Motion in a Plane (2026-27 reprint) covers projectile motion in Section 3.9 and uniform circular motion in Section 3.10, where it derives the centripetal acceleration a = v²/R. NCERT does not use the phrase "radius of curvature" for a projectile; the problem joins those two sections. The JEE (Main) 2026 syllabus lists "projectile motion, uniform circular motion", and the NEET (UG) 2026 syllabus lists the same topics. Check your own exam year's syllabus.
If you need the range, height and time-of-flight formulas first, start with our guide to projectile range and maximum height.
Why does v²/a give a radius?
In uniform circular motion, an object moving at speed v on a circle of radius R has acceleration v²/R towards the centre. Turn that round: if you know the speed and the part of the acceleration that points sideways, at right angles to the velocity, the radius of the matching circle is
R = v²/aₙ
where aₙ is called the normal component of acceleration. The part of the acceleration along the velocity only changes the speed; it does not bend the path, so it plays no part in R.
For a projectile the only acceleration is g, straight down. If the velocity makes an angle φ with the horizontal, the component of g perpendicular to the velocity is aₙ = g cos φ. Since cos φ = vₓ/v and the horizontal velocity vₓ = u cos θ never changes, this also gives a neat general form:
R = v²/(g cos φ) = v³/(g u cos θ)
What is R at the top and at launch?
At the top, the velocity is horizontal, so φ = 0, aₙ = g and v = u cos θ. So R_top = (u cos θ)²/g.
At launch, φ = θ and v = u. So R_launch = u²/(g cos θ).
Divide the two and you get R_top/R_launch = cos³θ. Because cos θ is less than 1, the path always bends most sharply at the top, where the speed is lowest and all of g acts sideways.
Worked example: one throw, three points
A ball is thrown at 30 m/s at an angle θ with tan θ = 4/3 (about 53°), so sin θ = 0.8 and cos θ = 0.6. Take g = 10 m/s².
Components: uₓ = 30 × 0.6 = 18 m/s and u_y = 30 × 0.8 = 24 m/s.
At launch. R = u²/(g cos θ) = 900/(10 × 0.6) = 150 m.
At the top. Time to the top = 24/10 = 2.4 s. The velocity is 18 m/s horizontally, so R = 18²/10 = 324/10 = 32.4 m.
At t = 1.0 s. v_y = 24 − 10 × 1.0 = 14 m/s and vₓ = 18 m/s, so v² = 324 + 196 = 520 and v ≈ 22.8 m/s. The angle of the velocity has cos φ = 18/22.8 ≈ 0.789, so aₙ ≈ 7.89 m/s². Then R = 520/7.89 ≈ 65.9 m. Check with the general form: v³/(g uₓ) = 22.8³/(10 × 18) ≈ 65.9 m.
Check the ratio. R_top/R_launch = 32.4/150 = 0.216, and cos³θ = 0.6³ = 0.216.
The radius falls from 150 m to 65.9 m to 32.4 m on the way up, then rises again in the same pattern on the way down, because the path is symmetric.
Which mistakes cost marks?
Using R = v²/g everywhere. That is only right at the top.
Using the launch speed at the top. At the top the speed is u cos θ, not u.
Using sin θ instead of cos θ. The component of g perpendicular to the velocity is g cos φ. Draw the velocity and g, and find the angle between g and the normal line before you write anything.
Calling R the height. The radius of curvature at the top (32.4 m here) has nothing to do with the maximum height, which is 24²/(2 × 10) = 28.8 m for this throw.
Forgetting units. v² in m²/s² divided by m/s² gives metres. Do the unit check on every line.
Once the method is clear, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, useful when your normal component comes out wrong and you cannot see why.
A practice routine
Find R at t = 2.0 s for the worked example. Check: v_y = 4 m/s, v² = 340, and R = v³/(g uₓ) ≈ 34.8 m.
Show that a horizontal throw from a cliff at speed u has R = u²/g at the moment of release.
For a launch at 45° with u = 20 m/s, show that R_launch ≈ 56.6 m and R_top = 20 m.
Time yourself on three mixed projectile questions, one of which asks for R.
Before your next session, write R_top/R_launch = cos³θ from memory and derive it in four lines.
