Physics 1Vectors and Motion in Two Dimensions

Launched Horizontally

Horizontal motion stays steady while vertical motion is free fall, and both share the same time.

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Roll a marble off a table. While it flies, gravity pulls it down exactly as in Free Fall, and nothing pushes it sideways. Treat the two components of its motion separately, and the problem splits into two easy ones that share the same clock.

Key idea

For a horizontal launch from height $h$: horizontally, $v_x$ stays constant; vertically, it falls from rest with $a = -9.8\ \text{m/s}^2$. The time in the air, $t = \sqrt{2h/g}$, depends only on the height.

Why it works

Forces change motion only along their own direction. Gravity is vertical, so it changes $v_y$ and leaves $v_x$ alone. With no air resistance, the horizontal motion is steady and the vertical motion is a drop from rest, so a ball launched sideways and a ball dropped at the same moment from the same height hit the floor together.
Launched sideways at 15 m/s from 20 m, beside a ball dropped at the same moment. They stay level with each other all the way down.

Worked examples

Worked example: Off a table

A ball rolls off a table $1.2\ \text{m}$ high at $3\ \text{m/s}$. How far from the table does it land?
  1. Vertical: $t = \sqrt{\dfrac{2(1.2)}{9.8}} \approx 0.495\ \text{s}$.
  2. Horizontal: $x = v_x t = 3 \times 0.495 \approx 1.48\ \text{m}$.

Answer: about $1.48\ \text{m}$

Worked example: Hitting a target

From a ledge $4.9\ \text{m}$ high, how fast must you throw a ball horizontally to land $10\ \text{m}$ away?
  1. Fall time: $t = \sqrt{\dfrac{2(4.9)}{9.8}} = 1\ \text{s}$.
  2. It must cover $10\ \text{m}$ in $1\ \text{s}$: $v_x = 10\ \text{m/s}$.

Answer: $10\ \text{m/s}$

Worked example: Landing speed

For the table ball above, how fast is it moving when it lands?
  1. $v_x = 3\ \text{m/s}$ still. $v_y = gt \approx 9.8 \times 0.495 \approx 4.85\ \text{m/s}$ downward.
  2. Speed $= \sqrt{3^2 + 4.85^2} \approx 5.70\ \text{m/s}$.

Answer: about $5.70\ \text{m/s}$

Common mistake

Thinking a ball fired horizontally stays up longer than one dropped from the same height, as if its speed "holds it up". The horizontal velocity has no vertical part, so it can't slow the fall. Check: $t = \sqrt{2h/g}$ has no $v_x$ in it.
Try it in the Science Lab's Orbit Lab: launch fast enough sideways from a high mountain and the ground curves away as fast as the cannonball falls. That is an orbit, and it was Newton's own thought experiment.
Try it in the Science LabOrbit Lab

Practice

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Sources and further reading

  • OpenStax College Physics 2e, Section 3.4 Projectile Motion
  • OpenStax University Physics Volume 1, Section 4.3 Projectile Motion
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