B. Kinematics · Kinematics: Motion in One and Two Dimensions

Projectile Motion: Objects Launched Horizontally

B2.8 B3.3 B2.7About 70 minutes

Learning goals

I can describe a projectile's motion as two independent motions: constant velocity horizontally and constant acceleration vertically.
I can explain why a ball launched horizontally and a ball dropped at the same instant land at the same time.
I can solve problems for objects launched horizontally: time of flight, range and final velocity.

Words to know

projectile
An object moving through the air with only gravity affecting it (air resistance ignored).

trajectory
The path of a projectile. Without air resistance it is a parabola.

range (Δdx)
The horizontal displacement of a projectile from launch to landing.

time of flight
The time a projectile spends in the air.

independence of motion
The horizontal and vertical motions of a projectile do not affect each other; they share only the time.

A race you can't win

Hold one coin flat on the edge of a table and flick it sideways off the edge, while at the same instant a second coin just drops from the edge. Which one lands first? Most people guess the dropped coin. Listen carefully: you hear a single click. They land at the same time. This surprising result is the key to all projectile motion.

Two motions at once

A projectile is any object moving through the air with only gravity acting on it: a thrown ball, a ski jumper, water from a hose. Ignoring air resistance, its motion splits into two parts that do not affect each other:

Horizontal (x): no acceleration (ax = 0). The horizontal velocity vx stays constant the whole flight. So Δdx = vxΔt.
Vertical (y): constant acceleration ay = 9.8 m/s² [down], exactly like the free-fall lesson. All five equations of motion apply.

The only thing the two motions share is time. Whatever you learn about Δt from one direction, you can use in the other. The combination of steady sideways motion and steadily growing downward speed gives the curved path, a parabola.

Two balls released at the same instant, shown every 0.2 s (to scale). Horizontally, the launched ball moves equal distances in equal times. Vertically, it falls exactly like the dropped ball.

Launched horizontally

When a projectile is launched horizontally, its initial vertical velocity is zero: v₁y = 0. That makes the vertical motion the same as a dropped object, so:

Time of flight comes from the vertical motion: Δdy = ½ayΔt², so Δt = √(2Δdy ÷ ay).
Range comes from the horizontal motion: Δdx = vxΔt.
Final velocity combines vx (unchanged) and v₂y = ayΔt with the Pythagorean theorem; its angle below the horizontal is tan⁻¹(v₂y ÷ vx).

For vertical calculations in this lesson, take down as positive, so ay = +9.8 m/s².

Worked example

Example 1: A cannonball fired from a cliff

A cannonball is fired horizontally at 30.0 m/s from the top of a cliff 45.0 m above level ground. Ignoring air resistance, find (a) the time of flight, (b) the range, and (c) the velocity just before it lands.

Vertical (down +): v₁y = 0, ay = 9.8 m/s², Δdy = 45.0 m.
(a) 45.0 = ½(9.8)Δt², so Δt² = 9.18 and Δt ≈ 3.03 s.
Horizontal: vx = 30.0 m/s for the whole flight.
(b) Δdx = vxΔt = (30.0)(3.030) ≈ 90.9 m.
(c) v₂y = ayΔt = (9.8)(3.030) ≈ 29.7 m/s [down]. |v₂| = √(30.0² + 29.7²) ≈ 42.2 m/s. Angle below horizontal: tan⁻¹(29.7 ÷ 30.0) ≈ 44.7°.
So v₂ ≈ 42.2 m/s at 44.7° below the horizontal (in the direction it was fired).

(a) 3.03 s; (b) 90.9 m; (c) 42.2 m/s at 44.7° below the horizontal

Example 1 drawn to scale. The horizontal velocity arrow is the same at every point. The vertical velocity arrow grows steadily. Their sum (dashed) is the actual velocity, which points more and more steeply downward.

Worked example

Example 2: Finding a launch speed

A marble rolls off a table 0.90 m high and lands 1.2 m from the foot of the table. How fast was it rolling when it left the table?

Vertical (down +): 0.90 = ½(9.8)Δt², so Δt = √(1.80 ÷ 9.8) ≈ 0.429 s.
Horizontal: vx = Δdx ÷ Δt = 1.2 ÷ 0.429 ≈ 2.8 m/s.
Notice the strategy: we never used vx to find time; time came from the vertical motion only.

About 2.8 m/s

Worked example

Example 3: Will the snowboarder clear the gap?

A snowboarder leaves a horizontal ledge at 8.0 m/s. The landing slope begins 5.0 m beyond the ledge and 2.5 m lower. Does she reach the landing slope, and how fast is she moving when she has dropped 2.5 m?

Time to drop 2.5 m: Δt = √(2 × 2.5 ÷ 9.8) ≈ 0.714 s.
Horizontal distance in that time: Δdx = 8.0 × 0.714 ≈ 5.7 m. That is more than 5.0 m, so yes, she clears the gap.
v₂y = 9.8 × 0.714 = 7.0 m/s [down]. |v₂| = √(8.0² + 7.0²) ≈ 10.6 m/s.

Yes, she travels about 5.7 m horizontally; her speed is about 10.6 m/s.

Projectile checklist

Make two columns: horizontal and vertical.
Horizontal: ax = 0, so the only equation is Δdx = vxΔt.
Vertical: ay = 9.8 m/s² [down]; for a horizontal launch, v₁y = 0.
Find Δt first, usually from the vertical column.
Use Δt in the other column.
For a final velocity, combine the components, and give an angle.

Just for fun

I asked a projectile's horizontal velocity how it stays so calm during the whole fall.

It said, "Gravity? Not my department."

Hands-on activity

Hit the target

You need: a smooth table, a marble or small ball (or a toy car), a ruler or metre stick, masking tape, a sheet of paper, and a stopwatch or phone slow-motion video. Grown-up note: marbles are a choking hazard for young children and a slipping hazard on the floor; collect them right away.

Mark a 50 cm stretch on the table that ends at the table edge. Roll the marble along it at a steady, gentle speed and time it (or film it) to find vx. Repeat until your launch speed is consistent.
Measure the table height Δdy.
Predict the landing point: Δt = √(2Δdy ÷ 9.8), then Δdx = vxΔt. Tape a paper target with an X at that distance from the foot of the table (use a plumb line, such as a string with a key tied on, to find the point directly below the edge).
Roll the marble at the same speed and see how close it lands. Calculate the percent difference between predicted and actual range.

What to look for: with a 0.75 m table and vx = 1.0 m/s (50 cm in 0.50 s), Δt ≈ 0.39 s and the range is about 0.39 m. Landing within a few centimetres of the X is a good result. The biggest uncertainty is usually vx, because rolling speed is hard to keep constant.

Your turn

A ball rolls off a 1.25 m high counter at 4.0 m/s. How long is it in the air?

How far from the base of the counter does it land?

Find its velocity just before landing.

A stone thrown horizontally from a 20 m high bridge lands 25 m away. How fast was it thrown?

Two balls roll off the same table at 1 m/s and 3 m/s. Which lands first? Which lands farther away?

What is a projectile's horizontal acceleration (no air resistance)? Its vertical acceleration?

More practice: warm-up

What shape is a projectile's trajectory without air resistance?

For a horizontal launch, what is v₁y?

Which motion (horizontal or vertical) usually gives you the time of flight?

A ball has vx = 5.0 m/s and is in the air for 2.0 s. What is its range?

Does a projectile's horizontal velocity change during the flight (no air resistance)?

More practice: on your own

A tennis ball is thrown horizontally at 12 m/s toward a wall 3.0 m away. How far does it drop before it hits the wall?

A diver runs off a 10.0 m platform horizontally at 2.0 m/s. How far from the platform's edge does she enter the water?

From what height must a ball be launched horizontally to be in the air for 1.5 s?

A ball launched horizontally lands 6.0 m away after 0.80 s. Find vx and the launch height.

If the launch height is quadrupled (same vx), what happens to the time of flight and the range?

Challenge

A bush plane flies horizontally at 60 m/s, 500 m above flat ground, and releases a package. Ignoring air resistance (real supply drops use parachutes), how far ahead of the target must it be released? Where is the plane when the package lands?

At what time after launch is the cannonball in Example 1 moving at 45° below the horizontal?

A ball rolls off a 1.0 m table. Show that its range is proportional to its launch speed, and find the constant.

Check your understanding

Quiz

Ignoring air resistance, the horizontal velocity of a projectile:

A ball is rolled off a table at the same moment another ball is dropped from the same height. Which lands first?

A ball is launched horizontally from 4.9 m high. Its time of flight is:

The ball in the last question was launched at 6.0 m/s. Its range is:

In horizontal-launch problems, you usually find the time of flight from:

Just before landing, a horizontally launched projectile's velocity points:

For parents and teachers

Big idea: A projectile's horizontal motion (constant velocity) and vertical motion (constant acceleration) are independent and linked only by time (B3.3). Students solve horizontal-launch problems such as the cannonball from a cliff named in the curriculum (B2.8).

How to help: Try the two-coin demonstration from the hook together and listen for one click. Ask your teen to set up two columns (horizontal, vertical) before calculating anything.

Common difficulties:

Using the launch speed in the vertical equations (for a horizontal launch, v₁y = 0).
Thinking a faster ball falls more slowly.
Forgetting to combine components for the final velocity, or giving no angle.
Mixing up range (horizontal) and height (vertical).