2.019
s
30.289
m
19.809
m/s
24.848
m/s
52.87
deg
1.657
x
9.905
m/s
2.019
s
30.289
m
19.809
m/s
24.848
m/s
52.87
deg
1.657
x
9.905
m/s
The Horizontal Projectile Motion Calculator analyzes the trajectory of an object launched horizontally from a height—such as a ball rolling off a table, a package dropped from an aircraft, or water flowing from a horizontal pipe. In this special case, the initial vertical velocity is zero, simplifying the projectile equations while retaining the essential physics of two-dimensional motion.
When an object is launched horizontally from height $$h$$, the time to reach the ground depends only on the height and gravity: $$t = \sqrt{\frac{2h}{g}}$$ This is identical to the time for a free-falling object dropped from the same height—the horizontal motion has no effect on the vertical fall.
The horizontal range is simply the horizontal velocity multiplied by the fall time: $$R = v_0 \sqrt{\frac{2h}{g}}$$ This equation reveals that range is proportional to $$v_0$$ (linear, not quadratic as in the angle-launched case) and proportional to the square root of height. Doubling the height increases range by a factor of $$\sqrt{2} \approx 1.414$$.
At impact, the projectile has two velocity components: the unchanged horizontal velocity $$v_x = v_0$$ and the acquired vertical velocity $$v_y = gt = \sqrt{2gh}$$. The total impact speed is: $$v_{\text{impact}} = \sqrt{v_0^2 + 2gh}$$ and the impact angle below the horizontal is: $$\alpha = \arctan\left(\frac{v_y}{v_x}\right) = \arctan\left(\frac{\sqrt{2gh}}{v_0}\right)$$
This configuration is especially important in forensic physics (determining launch speed from a fall), engineering (calculating where water jets or molten material will land), military applications (horizontal bombing runs), and physics education as the clearest demonstration that horizontal and vertical motions are independent.
The independence of horizontal and vertical motion is one of the most profound principles in classical mechanics, first demonstrated by Galileo. A bullet fired horizontally from a gun and a bullet dropped from the same height will hit the ground at exactly the same time—the horizontal velocity of the fired bullet does not affect its vertical acceleration.
Enter the drop height, horizontal velocity, and gravitational acceleration. The calculator computes fall time as $$t = \sqrt{2h/g}$$, horizontal range as $$R = v_0 t$$, vertical velocity at impact as $$v_y = gt$$, total impact speed as $$\sqrt{v_0^2 + v_y^2}$$, and the impact angle below horizontal using $$\arctan(v_y/v_0)$$.
Fall time depends only on height, not horizontal speed. Range scales linearly with horizontal velocity. The impact angle shows how steeply the object hits—a fast horizontal launch yields a shallow angle, while a slow launch from great height yields a steep angle. Impact speed always exceeds both the horizontal and vertical components individually.
Inputs
Results
The ball falls for 2.02 s, lands 30.3 m from the cliff base, and hits at 24.8 m/s at 52.9° below horizontal.
Inputs
Results
From 100 m at 50 m/s horizontal, the package lands 226 m ahead in 4.5 s.
No. Horizontal and vertical motions are independent. Fall time depends only on height and gravity: $$t = \sqrt{2h/g}$$. A faster horizontal launch means the object lands farther but at the same time.
The angle below the horizontal at which the object strikes the ground: $$\alpha = \arctan(v_y/v_x)$$. Slow horizontal launches from great heights yield steep angles near 90°; fast launches from low heights yield shallow angles.
Measure the horizontal distance $$R$$ and height $$h$$, then: $$v_0 = R / \sqrt{2h/g} = R\sqrt{g/(2h)}$$. This is commonly used in forensic physics to reconstruct accident scenarios.
Because the flight time is fixed by the height (not influenced by horizontal speed). In angled launches, increasing speed affects both the horizontal velocity and the flight time (via vertical component), giving a quadratic relationship.
No, this calculator assumes a purely horizontal launch (zero initial vertical velocity). For angled launches from a height, use the full oblique projectile equations with non-zero initial vertical velocity.
Range is proportional to $$\sqrt{h}$$. Doubling the height increases range by about 41% (factor of $$\sqrt{2}$$). Quadrupling the height doubles the range. This is weaker than the linear effect of horizontal velocity.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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