PhysicsCBSE Class 11 · Motion in a Plane

Projectile Motion Class 11: Time of Flight, Height and Range

Updated 17 October 2026 · Fact-checked

A projectile has constant horizontal velocity u cos θ and a vertical motion with acceleration −g. Time of flight T = 2u sin θ/g, maximum height H = u² sin²θ/2g and range R = u² sin 2θ/g, with the trajectory y = x tan θ − gx²/(2u² cos²θ), a parabola. Range is greatest at 45°, and θ and 90° − θ give equal ranges.

Understand Projectile Motion: Time, Height, Range and Trajectory

A projectile moves under gravity alone after being launched. Its two components are independent: in the horizontal direction there is no acceleration; in the vertical direction the acceleration is $- g$ .

**Launched at angle $θ$ with speed $u$ ** (level ground, $g = 10 m/s^{2}$ in the numericals below):

$u_{x} = u cos θ$ (constant), $u_{y} = u sin θ$ .
At time $t$ : $x = u cos θ t$ , $y = u sin θ t - \frac{1}{2} g t^{2}$ , $v_{x} = u cos θ$ , $v_{y} = u sin θ - g t$ .
Time to top $u sin θ / g$ ; time of flight $T = \frac{2 u sin θ}{g}$ .
Maximum height $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ .
Range $R = \frac{u ^{2} sin 2 θ}{g}$ , which is largest ( $u^{2} / g$ ) at $θ = 4 5^{\circ}$ , and equal for $θ$ and $9 0^{\circ} - θ$ .
Trajectory $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} cos ^{2} θ}$ , a parabola, and $tan θ = 4 H / R$ .

At the top the velocity is horizontal ( $u cos θ$ ), so kinetic energy is minimum there ( $\frac{1}{2} m u^{2} cos^{2} θ$ ); the acceleration is still $g$ downward.

**Horizontal projection from a height $h$ ** with speed $u$ : $t = 2 h / g$ , range $= u t$ , and on landing $v_{y} = g t$ , $v = u^{2} + v_{y}^{2}$ at angle $tan^{- 1} (v_{y} / u)$ below the horizontal.

Tips for harder items. To test whether a ball clears an obstacle, put the obstacle's horizontal distance into the trajectory equation and compare $y$ with its height. To catch a ball, compare the receiver's travel with the ball's range. Air resistance is ignored.

Projectile Motion: Time, Height, Range and Trajectory: formulas and key facts

Time of flight: $T = \frac{2 u sin θ}{g}$; Level ground.
Maximum height: $H = \frac{u ^{2} sin ^{2} θ}{2 g}$; At the top $v_{y} = 0$ .
Range: $R = \frac{u ^{2} sin 2 θ}{g}$; Maximum $u^{2} / g$ at 45°.
Trajectory: $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} cos ^{2} θ}$; Parabola.
Relation: $tan θ = \frac{4 H}{R}$; Eliminates $u$ .
Horizontal projection: $t = \frac{2 h}{g}, R = u t$; From height $h$ .

How to solve Projectile Motion: Time, Height, Range and Trajectory questions

Treat the two components separately and join them only through the time.

1Resolve the launch velocity into u cos θ and u sin θ.
2Write x(t) and y(t), or use the standard formulae T, H and R.
3For "when" questions, find t from the vertical motion.
4Use that time in the horizontal motion to get distance.
5For speed at an instant combine v_x and v_y: v = √(v_x² + v_y²) and tan φ = v_y/v_x.
6Check against R = v_x × T.

Projectile Motion: Time, Height, Range and Trajectory in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A goalkeeper's kick

Suppose a goalkeeper throws a ball at 18 m/s at 30° above the horizontal.

Find its time of flight and maximum height (g = 10 m/s²).

Try it first, then show the solutionSolution

u sin θ = 9 m/s; T = 2 × 9/10 = 1.8 s.
H = 9²/(2 × 10) = 4.05 m.

Answer: 1.8 s and 4.05 m.

Memory hook: Half the flight time up, half down.

A sprinkler jet

Suppose a garden hose is held horizontally at a height of 1.8 m and water leaves at 6 m/s.

How far from the foot of the hose does the water land?

Try it first, then show the solutionSolution

t = √(2 × 1.8/10) = 0.6 s.
Range = 6 × 0.6 = 3.6 m.

Answer: 3.6 m.

Memory hook: Time of fall decides the horizontal reach.

Tricks to remember Projectile Motion: Time, Height, Range and Trajectory

Short memory aids. Say each one aloud once, then use it on a question the same day.

Constant x, accelerated y
Horizontal: no acceleration. Vertical: −g.
Use it when: Setting up every problem.
Complementary angles
θ and 90° − θ give the same range.
Use it when: Range comparisons.
tan θ = 4H/R
Use it to find the angle from the shape of the path.
Use it when: Angle from H and R.
Top is minimum KE
Only the horizontal component survives at the top.
Use it when: Energy questions.

Projectile Motion: Time, Height, Range and Trajectory: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Projectile at 37° · 2 marks · Very short answerBoard level

A projectile is launched at 50 m/s at 37° above the horizontal (sin 37° = 0.6, cos 37° = 0.8, g = 10 m/s²). Find its time of flight, maximum height and range.

Try it first, then show the solutionSolution

u sin θ = 30, u cos θ = 40. T = 2 × 30/10 = 6 s; H = 900/20 = 45 m. (1)
R = u cos θ × T = 40 × 6 = 240 m (check: u² sin 2θ/g = 2500 × 0.96/10 = 240 m). (1)

Answer: T = 6 s; H = 45 m; R = 240 m.

Example 2: Package from an aircraft · 3 marks · Short answerBoard level

An aircraft flying horizontally at 80 m/s at a height of 500 m releases a package (g = 10 m/s²). Find the time to reach the ground, the horizontal distance travelled by the package, and its speed on landing.

Try it first, then show the solutionSolution

t = √(2 × 500/10) = 10 s. (1)
Horizontal distance = 80 × 10 = 800 m. (1)
vy = gt = 100 m/s; v = √(80² + 100²) = √16400 ≈ 128 m/s. (1)

Answer: 10 s; 800 m; about 128 m/s.

Example 3: Complementary angles · 3 marks · Short answerHarder board level

A ball is thrown at 30 m/s. Show that the ranges at 15° and 75° are equal, find the range, and compare the maximum heights.

Try it first, then show the solutionSolution

R = 900 × sin 30°/10 = 45 m at 15°; sin 150° = 0.5 gives the same 45 m at 75°. (1)
H(15°) = 900 sin²15°/20 ≈ 3.0 m; H(75°) = 900 sin²75°/20 ≈ 42.0 m. (1)
H₇₅/H₁₅ = tan²75° ≈ 13.9. (1)

Answer: R = 45 m for both; heights about 3.0 m and 42.0 m (ratio 13.9).

Example 4: Clearing a wall · 4 marks · Long answerHarder board level

A ball is kicked at 20 m/s at 45° from level ground (g = 10 m/s²). (a) Find its height when it is 20 m from the kicker horizontally. (b) Will it clear a 6 m wall placed 30 m away? (c) Find its range.

Try it first, then show the solutionSolution

y = x tan θ − gx²/(2u² cos²θ) with 2u² cos²θ = 2 × 400 × 0.5 = 400. (1)
(a) y(20) = 20 − 10 × 400/400 = 10 m. (1)
(b) y(30) = 30 − 10 × 900/400 = 7.5 m, which is above 6 m: it clears by 1.5 m. (1)
(c) R = u²/g = 400/10 = 40 m. (1)

Answer: (a) 10 m; (b) yes, by 1.5 m; (c) 40 m.

Example 5: Catching a kick · 4 marks · Long answerHOTS / competency

A player kicks a ball at 20 m/s at 37° above the horizontal (sin 37° = 0.6, cos 37° = 0.8; g = 10 m/s²). A teammate stands 50 m away and, at the instant of the kick, starts running towards the kicker at constant speed to receive the ball just as it lands. Find the ball's range, its time of flight and the speed of the runner.

Try it first, then show the solutionSolution

u sin θ = 12 m/s, u cos θ = 16 m/s. (1)
T = 2 × 12/10 = 2.4 s. (1)
R = 16 × 2.4 = 38.4 m. (1)
The runner covers 50 − 38.4 = 11.6 m in 2.4 s, so v = 11.6/2.4 ≈ 4.8 m/s. (1)

Answer: R = 38.4 m; T = 2.4 s; runner speed about 4.8 m/s.

Where students lose marks in Projectile Motion: Time, Height, Range and Trajectory

The slips examiners see most often, and what to write instead.

Mistake: Using the full speed in the horizontal motion.
Not resolving the launch velocity.
Do this: Use u cos θ horizontally and u sin θ vertically.
Usually costs: Method marks.
Mistake: Writing T = u sin θ/g for the full flight.
Forgetting the factor 2.
Do this: Time to the top is u sin θ/g; the flight is twice that.
Usually costs: 1 mark.
Mistake: Adding vx and vy as numbers for the speed.
Treating components as scalars.
Do this: v = √(vx² + vy²).
Usually costs: The answer mark.
Mistake: Saying the speed is zero at the top.
Confusing vy with the speed.
Do this: Only vy is zero; vx remains.
Usually costs: Reasoning mark.

Exam tips

State g and the origin; mention "air resistance neglected".
Show vx constant and vy changing as a two-line table in long answers.
For clearing-obstacle questions evaluate y at the obstacle's x and compare.
Use R = vx × T as a check on any range answer.
Check the angle convention (above or below the horizontal).

Practice questions on Projectile Motion: Time, Height, Range and Trajectory

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQHarder board level
A projectile is launched at 30° to the horizontal with kinetic energy K. Its kinetic energy at the top is
(a) K/4 (b) K/2 (c) 3K/4 (d) K
Hint
Only the horizontal component survives.
Answer
(c) 3K/4: K_top = K cos²30° = 3K/4.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Two balls thrown with the same speed at 20° and 70° to the horizontal have the same range.
Reason (R): The range is proportional to sin 2θ, and sin 40° = sin 140°.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
20° + 70° = 90°.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
A ball is rolled off the edge of a table 1.25 m high at 3 m/s. Find the time it takes to hit the floor and the horizontal distance from the table.
Hint
t = √(2h/g).
Answer
t = √(2 × 1.25/10) = 0.5 s; distance = 3 × 0.5 = 1.5 m.
Q4 · 3 marks · Short answerHarder board level
The maximum horizontal range of a projectile is 160 m. Find its speed of projection, the maximum height reached when it is launched at 45°, and the time of flight at 45° (g = 10 m/s²).
Hint
R_max = u²/g.
Answer
u² = gR = 1600, so u = 40 m/s; H = u² sin²45°/2g = 40 m; T = 2 × 40 × 0.707/10 ≈ 5.66 s.
Q5 · 4 marks · Long answerHOTS / competency
A projectile is launched at 60° above the horizontal with a speed of 40 m/s (g = 10 m/s²). Find its speed and direction of motion 2 s after launch, and say whether it is rising or falling.
Hint
Find vx and vy separately.
Answer
vx = 40 cos 60° = 20 m/s; vy = 40 sin 60° − 20 = 34.64 − 20 = 14.64 m/s. Speed = √(400 + 214.3) ≈ 24.8 m/s at tan⁻¹(14.64/20) ≈ 36° above the horizontal; vy > 0, so it is still rising.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Time of flight?Show answer
2u sin θ/g.
2.Maximum height?Show answer
u² sin²θ/2g.
3.Range?Show answer
u² sin 2θ/g.
4.Angle for maximum range?Show answer
45°.
5.tan θ in terms of H and R?Show answer
4H/R.
6.Path of a projectile?Show answer
A parabola.

Projectile Motion: Time, Height, Range and Trajectory: frequently asked questions

Why is the path a parabola?

Because x is proportional to t and y has a t² term, eliminating t gives y as a quadratic in x.

What is the speed at the top?

u cos θ, the horizontal component only.

Why do complementary angles give the same range?

Because R depends on sin 2θ and sin 2θ = sin(180° − 2θ).

What changes if the ground is not level?

The range and time of flight must be found from y(t) = target height, not the formulas for level ground.

What is the effect of air resistance?

It reduces the range and height; it is neglected at this level.