Drop Height of a Horizontally Launched Projectile

y=12gt2y = \tfrac{1}{2} g t^{2}

Worked example: 2 s of fall → 19.6133 m — press Try an example to run it live, then adjust anything.

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Drop Height of a Horizontally Launched Projectile explained

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Launch something horizontally and its sideways speed buys it no time at all: it falls exactly like a dropped object, y = ½gt², with g = 9.80665 m/s². After 2 s anything launched horizontally has fallen ½ × 9.80665 × 4 ≈ 19.6 m, whether it left the muzzle at 1 m/s or 800 m/s. Galileo argued this from his inclined-plane work; the modern demonstration is the classroom "monkey and hunter" apparatus, and MythBusters famously fired a bullet horizontally while dropping another and recorded both hitting the ground within milliseconds of each other.

Use it to time a fall from a table or a cliff, then multiply that time by the horizontal speed to get how far downrange the object lands. The trap is forgetting the ½ — an easy way to double every answer — and the second trap is a launch that is not truly horizontal, since any initial vertical component invalidates the simple square law. Real bullets deviate slightly because drag has a small vertical bite once they slow, but over short falls the agreement is startling.

Drop Height of a Horizontally Launched Projectile formula

y=12gt2y = \tfrac{1}{2} g t^{2}
Where
  • yy= Drop distance (m)
  • tt= Time of fall (s)

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