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Two balls are thrown at the same moment from different places. Will they hit each other in the air? Instead of writing four equations of motion and hoping the x and y values match, watch one ball from the other. This post gives the two conditions for a collision and works three cases with full numbers.
Is this on the syllabus?
The Syllabus for JEE (Main) 2026 lists the frame of reference, relative velocity and projectile motion under Unit 2, Kinematics. The NEET (UG) 2026 syllabus lists relative velocity and projectile motion too, and the JEE (Advanced) 2026 brochure lists kinematics in one and two dimensions and projectiles.
In the current NCERT Class 11 book, relative velocity is section 2.5 of Motion in a Straight Line, and Motion in a Plane covers projectiles in section 3.9. That chapter has no section on relative velocity in two dimensions; its points to ponder only remind you that v₁₂ = v₁ − v₂. So you build this method from NCERT pieces. Check the syllabus for your own exam year.
Why does the relative velocity stay constant?
Ignore air resistance. Both projectiles accelerate at g downward, so the acceleration of B relative to A is g − g = 0. With zero relative acceleration, the relative velocity u_B − u_A never changes. In A's frame, B moves in a straight line at constant speed from where it started. Gravity drops out completely.
That gives two conditions for a collision when both are launched together:
The relative velocity of B with respect to A must point from B's starting point straight at A's starting point.
The meeting time, separation ÷ relative speed, must come before either projectile hits the ground.
Worked example 1: two throws from level ground
A is at the origin and B is 60 m away on the same level ground. Take g = 10 m/s², sin 53° = 0.8 and cos 53° = 0.6. At the same instant, A throws a ball towards B at 25 m/s at 53°, and B throws one towards A at the same speed and angle.
Components: u_A = (15, 20) m/s and u_B = (−15, 20) m/s.
Relative velocity of B with respect to A: (−15 − 15, 20 − 20) = (−30, 0) m/s. It is horizontal and points straight at A, so condition 1 holds.
Meeting time: 60 ÷ 30 = 2.0 s.
Each ball's flight time is 2 × 20 ÷ 10 = 4.0 s, so both are still in the air. Condition 2 holds.
Ground-frame check at t = 2.0 s: A is at x = 15 × 2 = 30 m, y = 20 × 2 − 5 × 2² = 20 m. B is at x = 60 − 30 = 30 m, y = 20 m. They meet at (30 m, 20 m).
The general rule for two launches from the same level at the same instant: they can collide only if their vertical launch components are equal.
Worked example 2: a near miss
Same set-up, but B's ball leaves with components (−15, 15) m/s, about 21.2 m/s at 45°.
Relative velocity: (−15 − 15, 15 − 20) = (−30, −5) m/s, with magnitude √925 ≈ 30.4 m/s. It has a downward part, so B's straight line in A's frame passes below A. No collision.
Closest approach: d = |r₀ × v| ÷ |v| = |60 × (−5)| ÷ 30.4 ≈ 9.86 m.
It happens at t = −(r₀ · v) ÷ |v|² = 1800 ÷ 925 ≈ 1.95 s, before B's ball lands at 3.0 s.
Our post on closest approach of two moving bodies explains this formula.
Worked example 3: a ball thrown from a tower
B stands on a 40 m tower whose base is 40 m from A. At the same instant B throws a ball horizontally towards A at 5 m/s, and A launches a ball at (15, 20) m/s.
The line from A to B is (40, 40) m, at 45°.
Velocity of A's ball relative to B's: (15 − (−5), 20 − 0) = (20, 20) m/s, also at 45° and pointing at B. Condition 1 holds.
Distance √(40² + 40²) ≈ 56.6 m; relative speed √800 ≈ 28.3 m/s; meeting time 2.0 s.
B's ball would reach the ground at √(2 × 40 ÷ 10) ≈ 2.83 s, so the meeting comes first.
Check at t = 2.0 s: A's ball is at x = 15 × 2 = 30 m, y = 20 × 2 − 5 × 2² = 20 m. B's ball is at x = 40 − 5 × 2 = 30 m, y = 40 − 5 × 2² = 20 m.
If B simply drops the ball, A must aim straight at B's starting point: the classic monkey-and-hunter problem in one line.
Which mistakes cost marks?
Matching x and y at different times. A collision needs the same position at the same instant.
Skipping condition 2. A path that crosses after one ball has landed is not a collision.
Different launch times. Start the relative-motion clock at the second launch, using the first ball's position and velocity at that moment.
Adding speeds instead of subtracting vectors.
Why does switching frames feel strange?
In a study of 111 physics undergraduates from Bombay colleges, Panse, Ramadas and Kumar (1994) found that students tended to tie a frame of reference to a concrete object and to treat some motions as "real" and others as "apparent". That habit makes "B moves in a straight line as seen from A" feel like a trick. It is as valid as the ground's view, as each ground-frame check shows.
If you set up the relative velocity and still get a mismatch, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
A practice routine
Day 1: Redo example 1 with B throwing at 25 m/s at 37° (components −20 and 15 m/s). Show that the relative velocity (−35, −5) m/s means a miss.
Day 2: Three closest-approach problems, each checked from the ground.
Day 3: Two problems where the second ball is thrown a second later.
Day 4: One tower problem, checked from the ground.
Tonight, lower the tower in example 3 to 30 m and find B's throwing speed for a hit. You should get about 11.7 m/s, with the meeting at 1.5 s.
