Horizontal launch off a bluff: fall time, range, impact speed

A tennis-ball launcher is set up on a viewing platform at the brink of a 42.0 m lakeside bluff, its barrel clamped dead level so the launch is perfectly horizontal. Below, the lake surface sits level with the base of the bluff and runs flat out toward the horizon. The launcher is fired horizontally out over the water at 64.8 km/h: watchers on the platform see the ball leave on a flat line, bend away downward in an ever-steepening arc, and end in a white splash far out from the rock face. Air resistance is negligible. From the height and the muzzle speed alone, the whole flight — how long it lasts, how far it reaches, how hard it strikes the water — is already decided at the moment of firing.

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given

h = 42 m — Height of the bluff
v₀ = 64.8 km/h — Launch speed (horizontal)

Determine

(a)how long the ball is in the air
(b)how far from the base of the bluff it lands
(c)its vertical speed just before the water
(d)its impact speed

Step 1 of 4(a) · solve for Time of fall

The fall time comes from the height ALONE: y = ½gt², no launch speed anywhere in it. Fired at 64.8 km/h or simply dropped, the ball reaches the water at the same instant — Galileo's independence of motions, and the single fact this whole problem turns on.

Rearranged for t

t = \sqrt{\frac{2 y}{g}}

y

Your values, in your units

t = \sqrt{\frac{2 \, \left( 42\ \text{m} \right)}{\left( 9.80665\ \text{m/s}^{2} \right)}}

Answer

t = 2.9267\ \text{s}

in

Carried onward at full precision, not this rounded figure.

Open the Drop Height of a Horizontally Launched Projectile solver →

Step 2 of 4(b) · solve for Distance

Horizontally nothing pushes or drags, so the ball just cruises: 64.8 km/h becomes 18.0 m/s, held for the 2.93 s that gravity allows. Range is that simple product — the only way to land farther out is to fire faster or fall longer.

Rearranged for d

d = v t

v

t2.9267 scarried from step 1

Your values, in your units

d = \left( 64.8\ \text{km/h} \right) \, \left( 2.92671\ \text{s} \right)

Converted to base units

d = \left( 18\ \text{m/s} \right) \, \left( 2.92671\ \text{s} \right)

Answer

d = 52.681\ \text{m}

in

Carried onward at full precision, not this rounded figure.

Open the Speed, Distance & Time solver →

Step 3 of 4(c) · solve for Final velocity

The vertical story starts from zero — the launch was horizontal — and grows at g for the whole fall: v_y = gt. By the water it is 28.7 m/s, already half again larger than the horizontal 18.0 the ball has carried unchanged since the muzzle.

Rearranged for v

v = v_0 + a t

v0

a

t2.9267 scarried from step 1

Your values, in your units

v = \left( 0\ \text{m/s} \right) + \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 2.92671\ \text{s} \right)

Answer

v = 28.701\ \text{m/s}

in

Carried onward at full precision, not this rounded figure.

Open the Final Velocity (Uniform Acceleration) solver →

Step 4 of 4(d) · solve for Final velocity

Impact speed in one line: v² = v₀² + 2gh, which is energy conservation wearing kinematic clothes. It must agree with the Pythagorean sum of 18.0 across and 28.7 down — two routes, one 33.9 m/s, and the answer is a speed, never "18 + 28.7".

Rearranged for v

v = \sqrt{v_0^2 + 2 a d}

v0

a

d

Your values, in your units

v = \sqrt{\left( 64.8\ \text{km/h} \right)^2 + 2 \, \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 42\ \text{m} \right)}

Converted to base units

v = \sqrt{\left( 18\ \text{m/s} \right)^2 + 2 \, \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 42\ \text{m} \right)}

Answer

v = 33.879\ \text{m/s}

in

Carried onward at full precision, not this rounded figure.

Open the Velocity-Displacement Relation (v² = v₀² + 2ad) solver →

Why this order

Every horizontal-launch problem is two one-dimensional problems that happen to share a clock, and the chain's order makes the clock the first thing settled. Vertically the ball is in pure free fall from rest, so the height fixes the time; horizontally it is in pure constant-velocity motion, so that time fixes the range. The two calculations never borrow each other's numbers — the launch speed is absent from step 1 and g is absent from step 2 — and students who let them mingle (a 64.8 anywhere in the fall time, a ½gt² in the range) have missed the deepest fact in projectile motion: gravity cannot see sideways speed. Doubling the launch speed doubles the range and changes the flight time not at all.

Part (d) is where components are reassembled, and it offers a genuine choice of route. The chain uses v² = v₀² + 2gh, which looks like a kinematics equation but is really ½mv² = ½mv₀² + mgh with the mass cancelled — energy conservation, indifferent to the shape of the path. The other route is the Pythagorean sum of the two components from parts (b) and (c), √(18.0² + 28.7²), and the agreement of the two at 33.9 m/s is the self-check built into the problem. The error to retire permanently is adding the components like scalars: 18.0 + 28.7 = 46.7 m/s overstates the impact by 38%, because velocities at right angles combine as the diagonal of a rectangle, not as a sum. A speed is the LENGTH of the velocity arrow — that sentence, applied consistently, is worth more marks than any formula on this page.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.