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Projectile Motion Calculator — Range, Height & Time of Flight

Enter initial speed and launch angle to see the parabolic trajectory, key formulas, and step-by-step solutions. Ideal for Class 9–12 mechanics, JEE/NEET prep, and first-year university physics.

Parameters

m/sⓘ
°ⓘ
Show Trail

Controls

xⓘ

Calculated Values

Time of Flight:
3.60;s3.60;s
Maximum Height:
15.93;m15.93;m
Range:
63.71;m63.71;m
Horizontal Velocity:
17.68;m/s17.68;m/s
Initial Vertical Velocity:
17.68;m/s17.68;m/s

Examples

Example 1: Football kick (Class 11 style)

A ball is kicked from ground level at 20 m/s and 30° above the horizontal. Find time of flight, maximum height, and horizontal range. (Use g = 9.81 m/s².)

  • Time of Flight: 2.042.04
  • Maximum Height: 5.105.10
  • Range: 35.3635.36

Example 2: Stone thrown upward at an angle

A stone leaves a slingshot at 12 m/s and 50°. How long is it in the air, and how high does it rise before falling back to the same level?

  • Time of Flight: 1.881.88
  • Maximum Height: 6.776.77

Example 3: Fountain jet — range only

Water leaves a nozzle at 8 m/s at 60° to the horizontal. Assuming it lands at the same height, what is the horizontal range?

  • Range: 5.625.62

Example 4: Why 45° gives maximum range

Two launches use the same speed v₀ = 25 m/s but angles 35° and 55°. Because sin(2θ) is equal for supplementary pairs, both give nearly the same range (~51 m)—a common exam trap.

  • Range: 63.6663.66
  • Maximum Height: 15.9215.92
  • Time of Flight: 3.603.60

Visualization

What Is Projectile Motion?

Projectile motion describes the path of an object launched into the air and moving only under gravity (air resistance neglected). Near Earth's surface, gravity pulls the object downward with acceleration g ≈ 9.81 m/s², while horizontal motion stays uniform if no horizontal force acts on the body.

The motion splits into two independent parts: horizontal motion at constant speed v₀ cos θ, and vertical motion with initial speed v₀ sin θ and constant downward acceleration −g. Because horizontal velocity does not change, and vertical velocity changes linearly with time, the combined path is a parabola.

At the highest point (apex), vertical velocity is zero but horizontal velocity is unchanged. Total time in the air depends mainly on the vertical component of launch speed. Range—the horizontal distance from launch to landing on level ground—grows with both speed and angle, but for a fixed speed on flat ground the maximum range occurs at θ = 45°.

Use this calculator to check homework answers, explore how launch angle affects range and height, and connect graphs to the standard equations used in NCERT, CBSE, ICSE, and introductory university mechanics courses.

Key Concepts

  • Initial velocity (v₀): speed at launch; resolve into v₀ cos θ (horizontal) and v₀ sin θ (vertical)
  • Launch angle (θ): angle above the horizontal; 45° gives maximum range on level ground for a given speed
  • Time of flight (T): T = 2v₀ sin θ / g when launch and landing heights are equal
  • Maximum height (H): H = (v₀² sin² θ) / (2g), reached when vertical velocity becomes zero
  • Range (R): R = (v₀² sin 2θ) / g on level ground; depends on sin 2θ, so complementary angles (e.g. 30° and 60°) can share the same range
  • Assumptions: point mass, uniform gravity, no air drag—valid for many textbook and exam problems

Real-World Applications

  • Class 9–12 & board exams: numericals on football kicks, stone throws, and water jets from fountains
  • Sports science: optimizing kick/throw angle for distance or clearance over a barrier
  • Engineering: sprinkler design, firefighting hose reach, and fireworks shell paths (simplified models)
  • Military & aerospace: introductory ballistic estimates before drag and spin are included
  • Competitive exams (JEE/NEET): quick checks for 2D kinematics and symmetry of projectile paths

Explore Further

More mechanics tools

Physics Equations

Horizontal Position:
x(t)=v0cos⁡(θ)⋅tx(t) = v_0 \cos(\theta) \cdot t
Vertical Position:
y(t)=v0sin⁡(θ)⋅t−12gt2y(t) = v_0 \sin(\theta) \cdot t - \frac{1}{2}gt^2
Range:
Range=v02sin⁡(2θ)g\text{Range} = \frac{v_0^2 \sin(2\theta)}{g}
Maximum Height:
Max Height=v02sin⁡2(θ)2g\text{Max Height} = \frac{v_0^2 \sin^2(\theta)}{2g}
Time of Flight:
Time of Flight=2v0sin⁡(θ)g\text{Time of Flight} = \frac{2v_0 \sin(\theta)}{g}

Step-by-Step Solution

See how the main results are calculated.

1

Resolve Initial Velocity

Break initial velocity into horizontal and vertical components

Equation:

vx=v0cos⁡(θ),vy=v0sin⁡(θ)v_x = v_0 \cos(\theta), \quad v_y = v_0 \sin(\theta)

Calculation:

vx=25cos⁡(45°)=17.68 m/sv_x = 25 \cos(45°) = 17.68 \text{ m/s}
vy=25sin⁡(45°)=17.68 m/sv_y = 25 \sin(45°) = 17.68 \text{ m/s}

Result:

vx=17.68 m/s,vy=17.68 m/sv_x = 17.68 \text{ m/s}, v_y = 17.68 \text{ m/s}

Explanation:

The initial velocity is resolved into horizontal and vertical components using trigonometry.

2

Calculate Time to Peak

Find time to reach maximum height

Equation:

tpeak=vygt_{peak} = \frac{v_y}{g}

Calculation:

tpeak=17.689.81=1.80 st_{peak} = \frac{17.68}{9.81} = 1.80 \text{ s}

Result:

tpeak=1.80 st_{peak} = 1.80 \text{ s}

Explanation:

At the peak, vertical velocity becomes zero. Time to peak is found by dividing initial vertical velocity by gravitational acceleration.

3

Calculate Maximum Height

Find the maximum height reached

Equation:

hmax=h0+vy22gh_{max} = h_0 + \frac{v_y^2}{2g}

Calculation:

hmax=0+(17.68)22×9.81h_{max} = 0 + \frac{(17.68)^2}{2 \times 9.81}
hmax=0+15.93h_{max} = 0 + 15.93
hmax=15.93 mh_{max} = 15.93 \text{ m}

Result:

hmax=15.93 mh_{max} = 15.93 \text{ m}

Explanation:

Maximum height is calculated using the kinematic equation for vertical motion.

4

Calculate Range

Find the horizontal distance traveled

Equation:

R=vx×tflightR = v_x \times t_{flight}

Calculation:

R=17.68×3.60R = 17.68 \times 3.60
R=63.71 mR = 63.71 \text{ m}

Result:

R=63.71 mR = 63.71 \text{ m}

Explanation:

Range is the horizontal distance traveled, calculated by multiplying horizontal velocity by total flight time.

Frequently Asked Questions (FAQ)

What is projectile motion in physics?

Projectile motion is two-dimensional motion under uniform gravity, with no air resistance. The object has horizontal motion at constant velocity and vertical motion with constant acceleration g downward, producing a parabolic trajectory.

How do you calculate the range of a projectile?

On level ground, range R = (v₀² sin 2θ) / g, where v₀ is initial speed, θ is launch angle above horizontal, and g ≈ 9.81 m/s². This calculator applies that formula automatically from your inputs.

What is the formula for maximum height?

Maximum height H = (v₀² sin² θ) / (2g). It depends on the vertical component of launch speed. At the apex, vertical velocity is zero but horizontal velocity is still v₀ cos θ.

How do you find time of flight?

For launch and landing at the same height, time of flight T = 2v₀ sin θ / g. It depends on how long the vertical motion takes to rise and fall, not on horizontal speed.

Why is the path a parabola?

Horizontal displacement grows linearly with time (x = v₀ cos θ · t), while vertical displacement includes a t² term from gravity (y = v₀ sin θ · t − ½gt²). Eliminating t gives y proportional to x − (g/2v₀² cos² θ)x², which is a parabola.

What launch angle gives maximum range on flat ground?

For a fixed initial speed and level launch/landing, range is maximum at θ = 45° because sin 2θ reaches its maximum value of 1. Angles such as 30° and 60° give the same range for the same speed.

Does air resistance change projectile motion?

Textbook equations ignore drag. In real sports and ballistics, air resistance reduces range and height and can make the path non-parabolic. For school and most exam problems, the no-drag model is used.

Can I use these formulas on the Moon or other planets?

Yes. Replace g with the local gravitational acceleration (e.g. about 1.62 m/s² on the Moon). The same kinematic equations apply whenever gravity is approximately uniform over the flight.

Practice MCQs

  1. Which of the following best describes the path of a projectile (neglecting air resistance)?
  2. The horizontal component of velocity for a projectile (without air resistance) is:
  3. At the top of its trajectory, the vertical velocity of a projectile is:
  4. The time of flight of a projectile depends on:
  5. For a given speed, the maximum range is achieved at a launch angle of:

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