Horizontal Arrow Shot

An arrow is fired perfectly horizontally from a height $h$ at speed $v_0$. It starts with no vertical velocity, so it falls under gravity exactly the same as a dropped arrow — but moves sideways at $v_0$ the whole time. This is the cleanest demonstration of the independence of horizontal and vertical motion, and the answer often surprises students: a fast horizontal arrow falls in the same time as a slow one from the same height.

The Physics

With the launch point as origin, positive $x$ rightward and positive $y$ upward:

$x(t) = v_0\,t \qquad y(t) = h - \tfrac{1}{2}\,g\,t^{2}$

The arrow lands when $y = 0$. Solve for $t$, then plug back into $x$ for the range.

Step-by-Step Solution

Given:

• Arrow speed: $v_0 = 60\;\text{m/s}$
• Launch height: $h = 1.5\;\text{m}$
• Initial vertical speed: $v_{0y} = 0$ (horizontal launch)
• Gravity: $g = 9.81\;\text{m/s}^{2}$

Find: The horizontal distance traveled before the arrow hits the ground.

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Step 1 — Write the vertical position equation.

$y(t) = h - \tfrac{1}{2}\,g\,t^{2} = 1.5 - 4.905\,t^{2}$

Step 2 — Set $y = 0$ and solve for the landing time.

$1.5 - 4.905\,t^{2} = 0$

$t^{2} = \dfrac{1.5}{4.905} \approx 0.30581$

$t_{\text{land}} = \sqrt{0.30581} \approx 0.5530\;\text{s}$

Step 3 — Plug $t_{\text{land}}$ into the horizontal position equation.

The horizontal velocity is constant at $v_0 = 60\;\text{m/s}$:

$R = v_0\,t_{\text{land}} = 60 \times 0.5530 \approx 33.18\;\text{m}$

Step 4 — Find the impact velocity.

Horizontal: $v_x = 60\;\text{m/s}$ (unchanged).

Vertical (downward): $v_y = -g\,t_{\text{land}} = -(9.81)(0.5530) \approx -5.425\;\text{m/s}$

Impact speed:

$|\vec v| = \sqrt{v_x^{2} + v_y^{2}} = \sqrt{3600 + 29.43} = \sqrt{3629.43} \approx 60.24\;\text{m/s}$

The arrow only loses 0.24 m/s of speed from its 60 m/s launch — almost imperceptible — because gravity has so little time to act.

Step 5 — Find the impact angle below horizontal.

$\phi = \arctan\!\left(\dfrac{|v_y|}{v_x}\right) = \arctan\!\left(\dfrac{5.425}{60}\right) = \arctan(0.09042) \approx 5.17°$

Almost flat — the arrow strikes the ground at only 5° below horizontal.

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Answer: The arrow travels $R \approx 33.18\;\text{m}$ before hitting the ground, taking only $\approx 0.553\;\text{s}$ to fall the 1.5 m of release height. It strikes at $\approx 60.24\;\text{m/s}$ at an impact angle of just 5.2° below horizontal — so close to flat that it would barely scuff the dirt.

A Counter-Intuitive Truth

A dropped arrow and an arrow fired horizontally at any speed from the same height will hit the ground at the same instant. The forward velocity has no effect on how fast it falls — gravity acts purely on $v_y$. This is famously demonstrated by the bullet drop experiment (and by Mythbusters firing a bullet horizontally while dropping another). The visualization shows the dropped arrow as a small grey dot beside the archer — both reach the ground in the same number of frames.

Try It

• Increase the height — the arrow gets more flight time and travels much farther. At $h = 10\;\text{m}$, $R \approx 85.7\;\text{m}$.
• Increase the velocity — the arrow goes farther proportionally, but the flight time doesn't change.
• Watch the dropped arrow beside the archer: they always touch the ground simultaneously, regardless of the launch speed.