40.775
m
40.775
m
45
°
2.883
s
10.194
m
40.775
m
40.775
m
45
°
2.883
s
10.194
m
The Range of Projectile Calculator computes the horizontal distance a projectile travels when launched from and landing at the same height on level ground. This fundamental kinematics problem assumes no air resistance and constant gravitational acceleration, providing the idealized baseline for ballistic calculations.
The horizontal range is given by: $$R = \frac{v_0^2 \sin(2\theta)}{g}$$ where $$v_0$$ is the initial launch speed, $$\theta$$ is the launch angle measured from the horizontal, and $$g$$ is gravitational acceleration. The $$\sin(2\theta)$$ factor reveals the elegant angle-range relationship at the heart of projectile motion.
A remarkable property of projectile range is that complementary angles produce identical ranges. An object launched at 30° travels the same horizontal distance as one launched at 60° with the same speed, because $$\sin(2 \times 30°) = \sin(2 \times 60°) = \sin(60°) = \sin(120°)$$. This symmetry breaks down when air resistance or launch height differences are included.
Maximum range occurs at exactly 45°, where $$\sin(2\theta) = 1$$, giving $$R_{\max} = v_0^2 / g$$. This is a direct consequence of the equal partition of velocity between horizontal and vertical components. In practice, optimal launch angles are lower than 45° due to air resistance—competitive shot putters release at about 37°, and long jumpers take off at approximately 20°.
The range formula implicitly assumes the projectile lands at the same elevation it was launched from. When launch and landing heights differ, the range equation becomes more complex and the optimal angle shifts. Launching from elevation (like artillery on a hill) increases range, and the optimal angle becomes less than 45°. Conversely, launching uphill steepens the optimal angle.
Projectile range analysis has profound applications in sports science (optimizing throw distances), military ballistics, rocket trajectory planning, forensic analysis of falling objects, and even geological studies of volcanic ejecta. The calculator also shows the time of flight and maximum height to provide a complete picture of the trajectory.
Enter the initial velocity, launch angle, and gravitational acceleration. The calculator converts the angle to radians and applies $$R = v_0^2 \sin(2\theta) / g$$. It also computes the maximum possible range at 45° ($$v_0^2 / g$$), the complementary angle that gives the same range ($$90° - \theta$$), time of flight ($$T = 2v_0 \sin\theta / g$$), and maximum height ($$H = v_0^2 \sin^2\theta / (2g)$$).
The range is maximized at 45°. Complementary angles (e.g., 30° and 60°) yield equal ranges but different trajectories—higher angle means greater height and longer flight time but same horizontal distance. Range scales with the square of velocity, so doubling speed quadruples the range.
Inputs
Results
A ball kicked at 20 m/s at 45° travels 40.8 m—the maximum possible range for this speed.
Inputs
Results
At 30°, the javelin reaches 69.2 m—86.6% of the 79.9 m maximum range achievable at 45°.
At 45°, $$\sin(2\theta) = \sin(90°) = 1$$, its maximum value. This equally splits kinetic energy between horizontal and vertical components, optimizing the trade-off between flight time and horizontal speed.
Because $$\sin(2\theta) = \sin(180° - 2\theta) = \sin(2(90° - \theta))$$. The high-angle trajectory spends more time aloft but moves slower horizontally, while the low-angle trajectory is faster but shorter in flight time, balancing to equal range.
No. Air resistance makes the trajectory asymmetric, reduces range significantly (typically 30-60% for sports projectiles), and shifts the optimal angle below 45°. Numerical simulation is needed for realistic drag calculations.
When launching from height $$h$$ above the landing point, range increases and the optimal angle drops below 45°. The modified range requires solving a quadratic involving $$h$$, $$v_0$$, and $$\theta$$.
Range is inversely proportional to $$g$$. On the Moon ($$g = 1.62\,\text{m/s}^2$$), a projectile travels about 6 times farther than on Earth with the same initial conditions.
Range is proportional to $$v_0^2$$. Doubling the initial speed quadruples the range. This quadratic dependence makes speed far more impactful than angle adjustments for achieving greater distances.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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