Projectile motion for JEE Main

A projectile is simply two kinematics problems running at the same time: steady motion sideways and uniformly accelerated motion up and down. Once that picture is clear, every projectile formula follows in a line or two, and you stop needing to memorise them.

JEE Main questions on projectiles usually test the standard results with a twist: equal ranges, a trajectory equation to decode, a velocity at some instant, or a ball rolling off a table. This guide derives the results, lists the traps and works through problems of each type.

The assumptions

The standard results hold only when:

• air resistance is ignored;
• g is constant in magnitude and direction (small heights, flat ground);
• the launch and landing points are at the same level, unless stated otherwise.

If a question changes any of these, for instance by landing on a cliff or a slope, go back to the component equations instead of reaching for the formulas.

Splitting the motion

Launch a body with speed u at angle θ above the horizontal. Take x horizontal and y vertically up.

The horizontal velocity never changes. All the "interesting" physics is in the vertical direction.

The standard results, derived

Time of flight. The body lands when y = 0 again: (u sin θ)T = ½gT², so T = 2u sin θ / g.

Maximum height. At the top, v_y = 0. From v_y² = (u sin θ)² − 2gH: H = u² sin²θ / 2g. The time to reach the top is T/2.

Range. R = (u cos θ) × T = 2u² sin θ cos θ / g = u² sin 2θ / g.

Trajectory. Eliminate t using t = x / (u cos θ):
y = x tan θ − g x² / (2u² cos²θ), a parabola.
A neat alternative form is y = x tan θ (1 − x/R).

Complementary angles. Since sin 2θ = sin(180° − 2θ), the angles θ and (90° − θ) give equal ranges. The higher angle gives a greater height and a longer flight.

Projection from a height

A body is thrown horizontally at speed u from a height h.

• Time to land: t = √(2h/g). This is the same as dropping it, because the horizontal velocity does not affect vertical motion.
• Horizontal distance: x = u √(2h/g).
• Velocity on landing: vₓ = u, v_y = gt = √(2gh). Speed = √(u² + 2gh), at an angle tan⁻¹(v_y/vₓ) below the horizontal.

If the throw is at an angle from a height, write y = (u sin θ)t − ½gt² and set y = −h. You get a quadratic in t; take the positive root.

Projection on an inclined plane

Occasionally a question launches a projectile up a slope. Take axes along and perpendicular to the incline (angle β), with the launch at angle α to the incline. Gravity then has components g sin β along the slope and g cos β into it.

• Time of flight: T = 2u sin α / (g cos β).
• Maximum range up the incline: u² / (g(1 + sin β)).

With β = 0, these reduce to the ordinary results, which is a quick check.

Common mistakes

• Taking velocity at the top as zero. Only the vertical component is zero.
• Using R = u² sin 2θ / g for a ball landing below the launch point. That formula needs equal launch and landing heights.
• Mixing up the angle. Some questions give the angle with the vertical. Convert it first.
• Forgetting that 45° gives maximum range only on level ground and with no air resistance.

Worked problems

Problem 1: the standard set. A ball is projected at 40 m/s at 30° to the horizontal (g = 10 m/s²). Find T, H and R.

T = 2 × 40 × 0.5 / 10 = 4 s.
H = 40² × (0.5)² / 20 = 1600 × 0.25 / 20 = 20 m.
R = 1600 × sin 60° / 10 = 160 × (√3/2) = 80√3 ≈ 138.6 m.

Problem 2 (numerical answer): decoding a trajectory. A projectile follows y = x − x²/40 (SI units, g = 10 m/s²). Find its range and launch speed.

Comparing with y = x tan θ − g x² / (2u² cos²θ): tan θ = 1, so θ = 45°.
g / (2u² cos²θ) = 10 / (2u² × ½) = 10/u² = 1/40, so u² = 400 and u = 20 m/s.
Range: y = 0 at x = 40, so R = 40 m. Check: u²/g = 400/10 = 40.

Problem 3: ball off a table. A ball rolls off a table 20 m high at 5 m/s. Find where and how fast it lands.

t = √(2 × 20 / 10) = 2 s. Horizontal distance = 5 × 2 = 10 m.
v_y = 10 × 2 = 20 m/s, so speed = √(25 + 400) = √425 ≈ 20.6 m/s, at tan⁻¹(4) ≈ 76° below the horizontal.

Problem 4: velocity perpendicular to launch. A particle is projected at 20 m/s at 60°. When is its velocity perpendicular to the initial velocity?

The dot product of initial and current velocity must be zero:
(u cos θ)(u cos θ) + (u sin θ)(u sin θ − gt) = 0, which gives u² = ugt sin θ, so t = u / (g sin θ).
t = 20 / (10 × √3/2) = 4/√3 ≈ 2.31 s.
This is before landing (T = 2 × 20 × 0.866 / 10 ≈ 3.46 s), so it does happen. In general it can happen in flight only if θ > 45°; at exactly 45° it happens at the instant of landing.

Practice set

1. At what angle of projection is the range equal to the maximum height?
2. The maximum range of a gun on level ground is 100 m. What is the greatest height the shell reaches when fired for that range?
3. A body is projected at 60° with kinetic energy K. What is its kinetic energy at the highest point?
4. Two bodies are projected at 30° and 60° with the same speed. Find the ratio of their ranges and of their maximum heights.
5. A stone is thrown horizontally at 20 m/s from a cliff 45 m high. How long does it take to land, and how far from the base?
6. The speed of a projectile at its highest point is half its launch speed. What was the angle of projection?
7. A trajectory is y = 2x − 0.1x² (SI units). Find the range and the maximum height.

Answers

1. 4 cot θ = 1, so tan θ = 4, about 76°.
2. At 45°, R = u²/g = 100 m and H = u²/4g = 25 m.
3. K cos²60° = K/4.
4. Ranges 1 : 1 (complementary angles). Heights: sin²30° : sin²60° = 1 : 3.
5. t = √(90/10) = 3 s; distance = 20 × 3 = 60 m.
6. u cos θ = u/2, so θ = 60°.
7. y = 0 at x = 20 m (range). The top is at x = 10: y = 20 − 10 = 10 m. Check: R = 4H cot θ = 4 × 10 × ½ = 20.

What to do next

• Derive T, H, R and the trajectory equation from the component table without looking.
• Solve 25 mixed projectile problems, including five from a height.
• Revise kinematics if the vertical-motion steps feel slow.
• Next, take the same vector habits into circular motion.