PhysicsISC Class 11 · Motion in a Plane

Projectile Motion: Time of Flight, Height, Range and Trajectory

Updated 17 October 2026 · Fact-checked

A projectile moves under gravity alone. Its horizontal velocity stays constant and its vertical motion has acceleration g downward. For projection at speed u at angle θ: T = 2u sinθ/g, H = u² sin²θ/2g, R = u² sin2θ/g, and the path is a parabola.

Understand Projectile Motion Analysis

Independence of the two motions. Resolve the initial velocity into u cosθ (horizontal, constant) and u sinθ (vertical, decreasing at g). The two motions are independent and share only the time.

Formulae (level ground). Time of flight T = 2u sinθ/g. Maximum height H = u² sin²θ/(2g), reached at time T/2. Range R = u² sin2θ/g, which is greatest at θ = 45° and is the same for θ and 90° − θ. Also R = 4H/tanθ, so at 45° the range is four times the maximum height.

Trajectory. Eliminating t gives y = x tanθ − g x²/(2u² cos²θ). This is a parabola; comparing a given equation with this form gives θ and u. The range is the non-zero value of x for which y = 0.

Velocity at time t. vₓ = u cosθ and v_y = u sinθ − gt. The speed is √(vₓ² + v_y²) and the direction is tan⁻¹(v_y/vₓ). At the highest point the vertical velocity is zero and the speed is u cosθ, the minimum speed of the flight.

Horizontal projection from a height h. With speed u horizontally, time of fall T = √(2h/g), horizontal range = u√(2h/g), and on landing v_y = gT. Vertical and horizontal motions are as independent as before.

Assumptions. Air resistance is neglected, g is constant, and the Earth's curvature and rotation are ignored.

Projectile Motion Analysis: formulas and key facts

Level-ground projection: $T = 2 u s in θ / g ∣ H = u^{2} s i n^{2} θ /2 g ∣ R = u^{2} s in 2 θ / g$; u = speed of projection, θ = angle with the horizontal.
Trajectory: $y = x t an θ - g x^{2} / (2 u^{2} co s^{2} θ)$; A parabola through the point of projection.
Horizontal projection: $T = \sqrt (2 h / g) ∣ R = u \sqrt (2 h / g)$; h = height of the point of projection above the ground.
Range and height: $R = 4 H / t an θ$; At 45° the range is 4H.

Projectile Motion Analysis 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 stone thrown from a cliff

Suppose a boy standing on a cliff 80 m above a lake throws a stone horizontally at 15 m s⁻¹. Take g = 10 m s⁻².

Find the time to reach the water, the horizontal distance and the speed on impact.

Try it first, then show the solutionSolution

T = √(2h/g) = √(160/10) = 4 s.
Horizontal distance = 15 × 4 = 60 m.
Vertical speed on impact = gT = 40 m s⁻¹. Speed = √(15² + 40²) = √1825 = 42.7 m s⁻¹.

Answer: 4 s, 60 m, 42.7 m s⁻¹.

Memory hook: Fall time depends only on the height, not on the horizontal speed.

Tricks to remember Projectile Motion Analysis

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

Complementary angles, same range
sin2θ = sin(180° − 2θ), so θ and 90° − θ give equal ranges.
Use it when: Questions that give two angles of projection.
Top of the path
At the top v_y = 0, speed = u cosθ, acceleration still g.
Use it when: Velocity questions at maximum height.

Projectile Motion Analysis: solved examples

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

Example 1: Angle of maximum range · 1 mark · MCQBoard level

For a given speed of projection, the horizontal range is maximum when the angle of projection is:
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Try it first, then show the solutionSolution

R = u² sin2θ/g is greatest when sin2θ = 1, that is 2θ = 90°.

Answer: (b) 45°

Example 2: Time, height and range · 3 marks · Short answerBoard level

A projectile is fired at 30 m s⁻¹ at 30° above the horizontal. Find the time of flight, the maximum height and the range. (g = 10 m s⁻²)

Try it first, then show the solutionSolution

T = 2 × 30 × sin30°/10 = 2 × 30 × 0.5/10 = 3 s. (1 mark)
H = 900 × 0.25/20 = 11.25 m. (1 mark)
R = 900 × sin60°/10 = 90 × 0.866 = 77.9 m. (1 mark)

Answer: T = 3 s, H = 11.25 m, R = 77.9 m.

Example 3: Horizontal projection · 3 marks · Short answerHarder board level

A ball is thrown horizontally at 20 m s⁻¹ from the top of a 45 m tower. Find the time of flight, the horizontal range and the speed on reaching the ground. (g = 10 m s⁻²)

Try it first, then show the solutionSolution

T = √(2 × 45/10) = 3 s. (1 mark)
Range = 20 × 3 = 60 m. (1 mark)
v_y = 10 × 3 = 30 m s⁻¹; speed = √(400 + 900) = √1300 = 36.1 m s⁻¹. (1 mark)

Answer: 3 s, 60 m, 36.1 m s⁻¹.

Example 4: Complementary angles · 4 marks · Short answerHarder board level

A body is projected at 20 m s⁻¹ at (i) 30° and (ii) 60° to the horizontal. Show that the ranges are equal and compare the times of flight and maximum heights. (g = 10 m s⁻²)

Try it first, then show the solutionSolution

R = 400 sin60°/10 = 34.6 m for 30°, and 400 sin120°/10 = 34.6 m for 60°, since sin120° = sin60°. (1 mark)
T: 2 × 20 × 0.5/10 = 2 s for 30°; 2 × 20 × 0.866/10 = 3.46 s for 60°. (1 mark)
H: 400 × 0.25/20 = 5 m for 30°; 400 × 0.75/20 = 15 m for 60°. (1 mark)
The heights are in the ratio sin²30° : sin²60° = 1 : 3; the steeper throw stays longer and rises higher. (1 mark)

Answer: Equal range 34.6 m; T = 2 s and 3.46 s; H = 5 m and 15 m.

Example 5: Clearing a wall · 5 marks · Long answerHOTS / competency

A ball is projected at 25 m s⁻¹ at an angle with sinθ = 0.6 and cosθ = 0.8 (take these as exact, g = 10 m s⁻²). Find (i) the time of flight, maximum height and range, (ii) its height when 20 m and 40 m away horizontally, and (iii) the greatest height of a wall at 20 m that it can just clear.

Try it first, then show the solutionSolution

uₓ = 20 m s⁻¹, u_y = 15 m s⁻¹. T = 2 × 15/10 = 3 s; H = 225/20 = 11.25 m; R = 20 × 3 = 60 m. (2 marks)
x = 20 m at t = 1 s: y = 15 × 1 − 5 × 1 = 10 m. (1 mark)
x = 40 m at t = 2 s: y = 30 − 20 = 10 m, the same height by symmetry about the top at x = 30 m. (1 mark)
A wall at 20 m is cleared if it is under 10 m; the greatest height it can just clear is 10 m. (1 mark)

Answer: T = 3 s, H = 11.25 m, R = 60 m; height 10 m at both 20 m and 40 m; wall up to 10 m.

Where students lose marks in Projectile Motion Analysis

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

Mistake: Using the maximum-height formula for the range
Mixing up sin²θ/2g and sin2θ/g.
Do this: H uses sin²θ/(2g); R uses sin2θ/g.
Usually costs: 1 to 2 marks.
Mistake: Applying g to the horizontal motion
Writing x = u cosθ t − ½gt².
Do this: Horizontal acceleration is zero; g acts only on the vertical component.
Usually costs: 1 to 2 marks.
Mistake: Treating the speed at the top as zero
Only the vertical component is zero.
Do this: The speed at the highest point is u cosθ.
Usually costs: 1 mark.

Exam tips

Resolve u into components first and write them in the answer.
State g and the sign convention you use.
Check by symmetry: time to the top is T/2 and the speed at equal heights is the same.
For trajectory equations, compare coefficients with the standard form.

Practice questions on Projectile Motion Analysis

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 · MCQBoard level
A projectile is thrown at 20 m s⁻¹ at 45° (g = 10 m s⁻²). Its range is:
(a) 10 m
(b) 20 m
(c) 40 m
(d) 80 m
Hint
R = u² sin90°/g.
Answer
(c) 40 m
Q2 · 2 marks · Short answerBoard level
Show that for a projectile launched at 45° the range is four times the maximum height.
Hint
Write R and H at 45°.
Answer
R = u² sin90°/g = u²/g. H = u² sin²45°/2g = u²/4g. So R = 4H.
Q3 · 4 marks · Short answerHarder board level
A shell leaves the ground at 50 m s⁻¹ at an angle with sinθ = 0.8 and cosθ = 0.6 (g = 10 m s⁻²). Find the time of flight, maximum height, range and its velocity after 2 s.
Hint
uₓ = 30, u_y = 40.
Answer
T = 8 s; H = 1600/20 = 80 m; R = 30 × 8 = 240 m. At 2 s: v_y = 40 − 20 = 20 m s⁻¹, vₓ = 30, speed = √1300 = 36.1 m s⁻¹ at tan⁻¹(2/3) = 33.7° above the horizontal.
Q4 · 3 marks · Short answerHarder board level
A projectile follows y = 0.75x − 0.00625x² (SI units, g = 10 m s⁻²). Find the angle and speed of projection and the range.
Hint
Compare with y = x tanθ − g x²/(2u² cos²θ).
Answer
tanθ = 0.75 so cosθ = 0.8. 10/(2u² × 0.64) = 0.00625 → u² = 1250, u = 35.4 m s⁻¹. θ = 36.9°. Range: y = 0 at x = 0.75/0.00625 = 120 m.
Q5 · 3 marks · Short answerHarder board level
A pilot releases a food packet from a plane flying horizontally at 80 m s⁻¹ at a height of 500 m. How far ahead of the release point (horizontally) does the packet land, and what is its speed on landing? (g = 10 m s⁻²)
Hint
The packet starts with the plane's horizontal velocity.
Answer
T = √(1000/10) = 10 s. Range = 80 × 10 = 800 m. v_y = 100 m s⁻¹, speed = √(6400 + 10000) = √16400 = 128 m s⁻¹.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a cricketer throws a ball at 30 m s⁻¹ at 45° (g = 10 m s⁻²).
(i) Find the range. (1 mark)
(ii) Find the maximum height. (1 mark)
(iii) Find the speed at the highest point. (1 mark)
(iv) A fielder 60 m away runs to catch it at the same level; at what speed must he run to arrive exactly as the ball lands? (1 mark)
Hint
Time of flight is 2u sinθ/g.
Answer
(i) 900/10 = 90 m. (ii) 900 × 0.5/20 = 22.5 m. (iii) 30 cos45° = 21.2 m s⁻¹. (iv) T = 2 × 30 × 0.7071/10 = 4.24 s. He must run the extra 90 − 60 = 30 m away from the thrower in 4.24 s, which needs about 7.1 m s⁻¹.

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.Formula for the range?Show answer
R = u² sin2θ / g.
2.Angle for maximum range?Show answer
45°.
3.Speed at the top of the path?Show answer
u cosθ.
4.Shape of the trajectory?Show answer
A parabola.

Projectile Motion Analysis: frequently asked questions

Why is the horizontal velocity constant?

Gravity acts vertically, so there is no horizontal force (when air resistance is neglected), and by the first law the horizontal velocity stays unchanged.

Do two stones thrown horizontally at different speeds from the same height land together?

Yes. The fall time depends only on the height. The faster one lands farther away.

Is the range formula valid from a height?

No. R = u² sin2θ/g holds only when the landing and launching levels are the same. From a height, solve the vertical equation for the time first.