Horizontal Projectile from a Cliff

A ball thrown horizontally off a cliff is the simplest projectile motion case — and a perfect illustration of one of the most beautiful ideas in physics: horizontal and vertical motion are completely independent. Gravity pulls down the same way whether the ball is moving sideways or sitting still.

The Physics

Decompose the motion into two independent 1D problems. With the cliff edge as the origin, positive $x$ rightward and positive $y$ upward:

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

The horizontal velocity is constant (no horizontal force) and the vertical velocity grows linearly:

$v_x(t) = v_0 \qquad v_y(t) = -g\,t$

The strategy: solve $y(t) = 0$ for the landing time, then plug into $x(t)$ to get the range.

Step-by-Step Solution

Given:

• Cliff height: $h = 50\;\text{m}$
• Horizontal launch speed: $v_0 = 20\;\text{m/s}$
• Initial vertical speed: $v_{0y} = 0$ (thrown horizontally)
• Gravitational acceleration: $g = 9.81\;\text{m/s}^{2}$

Find: The horizontal landing distance and the time of flight.

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Step 1 — Write the vertical position equation with the given numbers.

Starting from $y(t) = h - \tfrac{1}{2}\,g\,t^{2}$ and substituting:

$y(t) = 50 - \tfrac{1}{2}(9.81)\,t^{2} = 50 - 4.905\,t^{2}$

Step 2 — Set $y = 0$ to find when the ball reaches the ground.

$50 - 4.905\,t_{\text{land}}^{2} = 0$

Move the $t^{2}$ term to the other side:

$4.905\,t_{\text{land}}^{2} = 50$

Divide both sides by $4.905$:

$t_{\text{land}}^{2} = \dfrac{50}{4.905} \approx 10.1937$

Take the positive square root:

$t_{\text{land}} = \sqrt{10.1937} \approx 3.193\;\text{s}$

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

The horizontal velocity is constant at $v_0 = 20\;\text{m/s}$, so:

$R = v_0 \cdot t_{\text{land}} = 20 \times 3.193 \approx 63.86\;\text{m}$

Step 4 — (Bonus) Find the impact velocity.

The horizontal component is unchanged: $v_x = 20\;\text{m/s}$. The vertical component has built up to:

$v_y = -g\,t_{\text{land}} = -(9.81)(3.193) \approx -31.32\;\text{m/s}$

The impact speed is the magnitude of the velocity vector:

$|\vec v| = \sqrt{v_x^{2} + v_y^{2}} = \sqrt{(20)^{2} + (31.32)^{2}} = \sqrt{400 + 980.95} \approx 37.16\;\text{m/s}$

And the impact angle below horizontal:

$\phi = \arctan\!\left(\dfrac{|v_y|}{v_x}\right) = \arctan\!\left(\dfrac{31.32}{20}\right) = \arctan(1.566) \approx 57.43°$

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Answer: The ball lands ≈ 63.86 m from the base of the cliff after a flight time of ≈ 3.19 s, striking the ground at about 37.2 m/s at an angle of 57.4° below horizontal.

Try It

• Increase the height — the ball stays in the air longer and lands proportionally farther. Notice $R \propto \sqrt{h}$, so doubling the height multiplies the range by $\sqrt{2} \approx 1.414$.
• Increase the launch velocity — the ball travels farther without changing how long it falls. The two are decoupled!
• Change gravity — see what would happen on the Moon or on Jupiter.
• Watch the cyan velocity arrow stay the same length while the pink velocity arrow grows steadily — that's gravity acting on $v_y$ alone.