JEE Mains Physics · Motion in a Plane

Projectile: Time of Flight, Maximum Height and Range

A projectile launched from level ground at speed u and angle θ stays in the air for 2u sin θ/g, rises to u² sin²θ/2g and lands u² sin 2θ/g away; the range is greatest at 45°, and angles θ and 90° − θ give the same range.

All Motion in a Plane notes JEE Mains Physics strategy

Why this matters

Twenty-seven PYQs, twenty-two of them multiple choice, and three from 2026. Fourteen use the three formulas directly, most often as a ratio between two launches; seven ask for the greatest range or for the angle at which the range is a set multiple of the height; six are about two launches at complementary angles. Mass never appears in any of the formulas, so a question that changes the mass and nothing else changes no answer.

Concept 1 of 3: Time of flight, maximum height and range

Split the launch velocity in two. The horizontal part u cos θ never changes, because nothing acts sideways. The vertical part u sin θ is thrown straight up against g. So the time in the air comes from the vertical part alone, the height from the vertical part alone, and the range is the horizontal speed times the time in the air.

Definition

$T = \dfrac{2u\sin\theta}{g}$ ; time to the top $T/2$ .
$H = \dfrac{u^{2}\sin^{2}\theta}{2g}$ .
$R = u\cos\theta \cdot T = \dfrac{u^{2}\sin 2\theta}{g}$ .
At a fixed speed: $T \propto \sin\theta$ , $H \propto \sin^{2}\theta$ , $R \propto \sin 2\theta$ . At a fixed angle, each of H and R goes as $u^{2}$ and T as u.
Same time of flight means the same vertical component $u\sin\theta$ , and so the same height, whatever the masses.
Angles $45^{\circ} \pm \alpha$ at the same speed: times compare as $\sin(45^{\circ} + \alpha) : \sin(45^{\circ} - \alpha)$ and heights as the squares; expand with $\sin(45^{\circ} \pm \alpha) = \dfrac{\cos\alpha \pm \sin\alpha}{\sqrt{2}}$ , and use $(\cos\alpha \pm \sin\alpha)^{2} = 1 \pm \sin 2\alpha$ for the squares.
The angle measured from the VERTICAL is $90^{\circ} - \theta$ : then $H = \dfrac{u^{2}\cos^{2}\phi}{2g}$ .

Projectile on level ground

T = \frac{2u\sin\theta}{g} \qquad H = \frac{u^{2}\sin^{2}\theta}{2g} \qquad R = \frac{u^{2}\sin 2\theta}{g}

Worked example

A ball is kicked at 20 m/s at

37^{\circ}

to the ground

(\sin 37^{\circ} = 0.6,\ \cos 37^{\circ} = 0.8,\ g = 10\ \text{m/s}^{2})

. Find its time of flight, greatest height and range.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q6Moderate

Example 1 · Motion in a Plane · Projectile: Time of Flight, Maximum Height and Range

The two projectiles are projected with the same initial velocities at the

15^{\circ}

and

30^{\circ}

with respect to the horizontal. The ratio of their ranges is

1:x

. The value of

x

sin 2θ in the range, sin²θ in the height

R = u² sin 2θ/g and H = u² sin²θ/2g. Swapping them is the commonest slip; check with 90°, where the range must be zero and the height u²/2g.

Time of flight does not use sin 2θ

T = 2u sin θ/g comes from the vertical motion alone. Ranges compare as sin 2θ, but times compare as sin θ and heights as sin²θ.

Drill 13 more on time of flight, maximum height and range

Concept 2 of 3: Maximum range and the range-height relation

sin 2θ cannot exceed 1, so the range at a given speed is largest when 2θ = 90°, that is at 45°. Thrown straight up, the same ball rises u²/2g: exactly half the greatest range. Dividing the range by the height cancels the speed and leaves only the angle, which is why 'range equals n times height' fixes θ.

Definition

$R_{\max} = \dfrac{u^{2}}{g}$ at $\theta = 45^{\circ}$ (on level ground).
Greatest height thrown vertically $= \dfrac{u^{2}}{2g} = \dfrac{R_{\max}}{2}$ .
$\dfrac{R}{H} = \dfrac{4}{\tan\theta}$ , so $R = nH$ gives $\tan\theta = \dfrac{4}{n}$ . For $\tan\theta = 4/n$ , $\sin 2\theta = \dfrac{8n}{n^{2} + 16}$ .
Time to the top t and range R: $u\sin\theta = gt$ and $R = 2u\cos\theta \cdot t$ , so $\cot\theta = \dfrac{R}{2gt^{2}}$ .

Greatest range; range against height

R_{\max} = \frac{u^{2}}{g}\ (45^{\circ}) \qquad \frac{R}{H} = \frac{4}{\tan\theta}

Worked example

A ball is thrown at 20 m/s so that its range is twice its greatest height

(g = 10\ \text{m/s}^{2})

. Find the angle of projection, the range and the height.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q86Moderate

Example 2 · Motion in a Plane · Projectile: Time of Flight, Maximum Height and Range

A particle is projected with velocity

u

so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as

\frac{nu^{2}}{25g}

, where value of

n

is : (Given '

g

' is the acceleration due to gravity).

Greatest height equal to greatest range

The greatest height (thrown straight up) is u²/2g; the greatest range (at 45°) is u²/g. The range is twice the height, not equal to it.

45° is only for level ground

Range is largest at 45° only when the ball lands at its launch height. Thrown from a tower or onto a slope, the best angle is different.

Drill 6 more on maximum range and the range-height relation

Concept 3 of 3: Complementary angles give the same range

sin 2θ = sin(180° − 2θ), so a launch at θ and a launch at 90° − θ, at the same speed, land at the same point. The steeper one goes higher and stays up longer; the flatter one is quicker and lower. Because sin(90° − θ) = cos θ, the two heights and the two times of flight combine into neat products that give the range directly.

Definition

Same speed, angles θ and $90^{\circ} - \theta$ : equal ranges, unequal heights and times.
$H_1 = \dfrac{u^{2}\sin^{2}\theta}{2g}$ , $H_2 = \dfrac{u^{2}\cos^{2}\theta}{2g}$ : so $H_1 + H_2 = \dfrac{u^{2}}{2g}$ and $R = 4\sqrt{H_1H_2}$ .
$T_1 = \dfrac{2u\sin\theta}{g}$ , $T_2 = \dfrac{2u\cos\theta}{g}$ : so $T_1T_2 = \dfrac{2R}{g}$ .
For any range below the greatest, exactly two angles reach it, and they add to $90^{\circ}$ .

Two angles, one range

R = 4\sqrt{H_1H_2} \qquad T_1T_2 = \frac{2R}{g} \qquad H_1 + H_2 = \frac{u^{2}}{2g}

Worked example

Two balls are thrown at 30 m/s and land at the same point; one is thrown at

37^{\circ}

(\sin 37^{\circ} = 0.6,\ g = 10\ \text{m/s}^{2})

. Find the other angle, the two greatest heights and the range.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q6Moderate

Example 3 · Motion in a Plane · Projectile: Time of Flight, Maximum Height and Range

Two identical bodies projected with the same speed at two different angles cover the same horizontal range R . If the time of flight of these bodies are 5 s and 10 s , respectively, then the value of

R

\_\_\_\_

m. (Take

g = 10\text{ }m/s^{2}

)

Equal range does not mean equal height

θ and 90° − θ land together, but the steeper launch rises higher and stays up longer. Only the range is shared.

Adding the angles to 180°

The two angles that give one range add to 90°, not 180°. It is the doubled angles, 2θ and 180° − 2θ, that add to 180°.

Drill 5 more on complementary angles give the same range

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Time of flight, maximum height and range
Projectile on level ground
$T = \frac{2u\sin\theta}{g} \qquad H = \frac{u^{2}\sin^{2}\theta}{2g} \qquad R = \frac{u^{2}\sin 2\theta}{g}$
Maximum range and the range-height relation
Greatest range; range against height
$R_{\max} = \frac{u^{2}}{g}\ (45^{\circ}) \qquad \frac{R}{H} = \frac{4}{\tan\theta}$
Complementary angles give the same range
Two angles, one range
$R = 4\sqrt{H_1H_2} \qquad T_1T_2 = \frac{2R}{g} \qquad H_1 + H_2 = \frac{u^{2}}{2g}$

Watch out for (6)

sin 2θ in the range, sin²θ in the height
→ Time of flight, maximum height and range
Time of flight does not use sin 2θ
→ Time of flight, maximum height and range
Greatest height equal to greatest range
→ Maximum range and the range-height relation
45° is only for level ground
→ Maximum range and the range-height relation
Equal range does not mean equal height
→ Complementary angles give the same range
Adding the angles to 180°
→ Complementary angles give the same range

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