B. Dynamics · Dynamics: Forces and Motion in a Plane

Projectile Motion: Any Launch Angle, Any Height

B2.2 B2.1 A1.5 A1.8About 75 minutes

Learning goals

I can split a projectile's motion into independent horizontal (constant velocity) and vertical (constant acceleration) motions.
I can solve projectile problems where the launch and landing heights are different, using the quadratic formula for time.
I can find a projectile's velocity (magnitude and direction) at any moment, including at impact.
I can derive the range equation R = v₁² sin 2θ ÷ g and explain when it may and may not be used.
I can plan and carry out a controlled test of a projectile prediction and compare the result with the prediction.

Words to know

projectile
An object moving through the air under the influence of gravity alone (air resistance ignored).

trajectory
The path a projectile follows. With no air resistance it is a parabola.

range
The horizontal displacement of a projectile when it lands.

time of flight
The total time a projectile is in the air. It is set by the vertical motion.

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

The snowball from the cliff

You stand on a rocky point 25 m above a frozen lake and throw a snowball up and out at 15 m/s. Does it go farther than if you stood at the lake's edge? How long is it in the air, and how fast is it going when it lands?

In SPH3U you solved projectiles launched horizontally and projectiles launched at an angle from level ground. Grade 12 removes the training wheels: any angle, any starting height, any target. The physics is the same; the algebra is a little richer.

Remember from Grade 11

A projectile has two independent motions that share one clock:

Horizontal: no force (air resistance ignored), so ax = 0 and vx stays constant: Δx = vx Δt.
Vertical: gravity only, so ay = −9.8 m/s² (up positive). Use the kinematics equations, for example Δy = v₁y Δt + ½ay Δt² and v₂y = v₁y + ay Δt.
Resolve the launch velocity: v₁x = v₁ cos θ, v₁y = v₁ sin θ.

The general method

Draw the path and choose axes. Put the origin at the launch point, with up positive.
Resolve the launch velocity into v₁x and v₁y.
Find the time from whichever direction gives enough information. Usually that is the vertical motion: Δy = v₁y Δt − 4.9 Δt². If the landing point is not at the launch height, this is a quadratic in Δt. Use the quadratic formula and keep the positive root.
Use that time in the other direction (Δx = v₁x Δt).
For the velocity at any moment: v₂x = v₁x and v₂y = v₁y − 9.8 Δt. Combine them with Pythagoras and tan⁻¹.

Notice that the time is the bridge between the two directions. Everything else is kept separate.

Worked example

Example 1: Launched above the landing point

A snowball is thrown at 15.0 m/s at 35° above the horizontal from a point 25.0 m above a frozen lake. Find (a) the time of flight, (b) how far from the base of the cliff it lands, and (c) its velocity at impact.

v₁x = 15.0 cos 35° = 12.29 m/s; v₁y = 15.0 sin 35° = 8.60 m/s.
(a) Vertical, up +: Δy = −25.0 m. So −25.0 = 8.60Δt − 4.9Δt², which rearranges to 4.9Δt² − 8.60Δt − 25.0 = 0.
Δt = [8.60 + √(8.60² + 4(4.9)(25.0))] ÷ (2 × 4.9) = [8.60 + √(74.0 + 490)] ÷ 9.8 = (8.60 + 23.75) ÷ 9.8 = 3.30 s. (The other root is negative and has no meaning here.)
(b) Δx = v₁x Δt = 12.29 × 3.30 = 40.6 m.
(c) v₂x = 12.29 m/s. v₂y = 8.60 − 9.8(3.30) = −23.7 m/s.
Speed = √(12.29² + 23.7²) = 26.7 m/s. Angle below horizontal = tan⁻¹(23.7 ÷ 12.29) = 62.6°.

(a) 3.30 s (b) 40.6 m (c) 26.7 m/s at 63° below the horizontal.

The snowball's path from Example 1 (not to scale). The horizontal velocity, 12.3 m/s, is the same at every point. Only the vertical velocity changes: 8.6 m/s up at launch, zero at the top, 23.7 m/s down at impact.

The range equation, and when not to use it

For the special case of landing at the launch height, you can derive a single formula. The time of flight comes from Δy = 0:

0 = v₁ sin θ Δt − ½gΔt², so Δt = 2v₁ sin θ ÷ g.

Then R = v₁x Δt = v₁ cos θ × 2v₁ sin θ ÷ g. Using the identity 2 sin θ cos θ = sin 2θ:

R = v₁² sin 2θ ÷ g (level ground only)

Two results fall straight out of it. sin 2θ is largest (equal to 1) when 2θ = 90°, so the maximum range is at 45°. And because sin(180° − 2θ) = sin 2θ, complementary angles such as 30° and 60° give the same range. The higher shot takes longer and goes higher, but lands at the same spot.

Warning: the range equation is wrong whenever the projectile lands higher or lower than it started. For the snowball in Example 1 it would predict 21.6 m instead of 40.6 m. When in doubt, use the general method.

Range against launch angle for a 20 m/s launch on level ground (no air resistance). The maximum is at 45°, and pairs of complementary angles such as 15° and 75° give equal ranges.

Worked example

Example 2: Using the range equation

A golfer wants a 150 m carry on a level fairway with a launch angle of 40°. Ignoring air resistance, what launch speed is needed?

Same launch and landing height, so the range equation applies.
R = v₁² sin 2θ ÷ g, so v₁ = √(Rg ÷ sin 2θ).
v₁ = √(150 × 9.8 ÷ sin 80°) = √(1470 ÷ 0.9848) = √1492.7 = 38.6 m/s.

About 39 m/s. Real golf balls need more, because air resistance shortens the flight (though backspin also creates lift).

Worked example

Example 3: Clearing the crossbar

A soccer player takes a free kick 25.0 m from the goal line. The ball leaves the ground at 22.0 m/s at 20.0° above the horizontal. The crossbar is 2.44 m high. Does the ball pass under the crossbar? Is it rising or falling as it crosses the goal line?

v₁x = 22.0 cos 20.0° = 20.67 m/s; v₁y = 22.0 sin 20.0° = 7.52 m/s.
This time the horizontal motion gives the time: Δt = 25.0 ÷ 20.67 = 1.209 s.
Height then: Δy = 7.52(1.209) − 4.9(1.209)² = 9.10 − 7.17 = 1.93 m.
Vertical velocity then: v₂y = 7.52 − 9.8(1.209) = −4.33 m/s (negative, so it is falling).

The ball is 1.93 m high, so it passes under the 2.44 m crossbar, and it is already falling (dipping) as it crosses the line.

Real projectiles

Air resistance makes real trajectories shorter and steeper on the way down than a parabola. The ideal model is still the starting point for engineers and sports scientists; they add drag afterwards. Also, when a projectile is launched from above its landing point (a shot put, a snowball from a cliff), the angle for maximum range is less than 45°.

Just for fun

In 1971, Apollo 14 astronaut Alan Shepard hit golf balls on the Moon. With lunar gravity only about one-sixth of Earth's and no air, a projectile launched the same way would travel about six times as far as on airless Earth (his bulky suit meant he could only swing one-handed, though).

Hands-on activity

Inquiry: predict where the marble lands

Question: Can you predict where a marble will land when it rolls off a table?

You need: a marble, a ramp (a ruler with a groove, or a folded file folder) propped on a book, a table, masking tape, a measuring tape, a sheet of paper and a pencil. Small marbles are a choking hazard: keep them away from young children and pick them all up afterwards.

Tape the ramp so its bottom end is a few centimetres from the table edge. Mark a release point on the ramp so every roll starts the same way (a controlled variable).
Find the launch speed: the marble leaves the table horizontally. Measure the table height h. Roll it 5 times and mark the landing spots on paper taped to the floor. Measure the average horizontal distance x. Then Δt = √(2h ÷ 9.8) and v = x ÷ Δt.
Predict: place the table (or the ramp on a sturdy box on the floor) at a new height h₂. Using the same v, predict the new landing distance x₂ = v√(2h₂ ÷ 9.8). Put a paper cup or a target mark at x₂.
Test: release from the same point 5 times. Measure the average distance and find the percent difference: |measured − predicted| ÷ predicted × 100%.
Analyse: name your independent, dependent and controlled variables, and two sources of error (for example, the marble spinning, release point, or measuring to the centre of the marks). Was your prediction within your uncertainty?

Your turn

A ball is kicked from level ground at 12 m/s at 50° above the horizontal. Find the time of flight, maximum height and range.

Use the range equation to find the range of a projectile launched at 25 m/s at 30° on level ground. What other angle gives the same range?

A stone is thrown horizontally at 8.0 m/s from a bridge 20 m above a river. How far from the bridge does it land, measured horizontally?

For the stone above, find its velocity just before it hits the water.

A ball is thrown from a height of 1.5 m at 10 m/s at 45°. How long is it in the air before it hits the (level) ground?

Why is the range equation not valid for the ball in the previous question?

More practice: warm-up

What is the horizontal acceleration of a projectile (no air resistance)?

What is the vertical acceleration of a projectile?

What is the vertical velocity at the top of the path?

Which launch angle gives the maximum range on level ground?

A projectile is launched at 20° on level ground. Which other angle gives the same range?

More practice: on your own

A cannonball is fired at 30 m/s at 60° above level ground. Find its maximum height, time of flight and range.

A ball is launched at 20 m/s at 40°. Find its velocity 1.0 s after launch.

A garden hose sprays water at 12 m/s from ground level. At what two angles must it be aimed to reach a flower bed 10 m away?

A ball is thrown from 2.0 m above level ground at 15 m/s at 30° above the horizontal. Find the horizontal distance it travels.

A student uses the range equation for the snowball in Example 1 and gets 21.6 m. Explain the error and the size of the mistake.

Challenge

Derive an expression for the maximum height H of a projectile launched with speed v₁ at angle θ on level ground. Then show that at θ = 45°, H is exactly one quarter of the range.

A snowball is thrown at 14 m/s at 45° toward a wall 12 m away. At what height does it hit the wall, and is it rising or falling then?

On the Moon, g = 1.62 m/s². If a ball is launched with the same velocity on the Moon and on (airless) Earth from level ground, how do the ranges compare?

Check your understanding

Quiz

At the highest point of its path, a projectile's acceleration is:

A ball is launched at 35° and another at 55°, both at the same speed on level ground. Compared with the first, the second ball:

The range equation R = v₁² sin 2θ ÷ g can be used when:

A projectile is launched at 20 m/s at 30° above the horizontal. Its horizontal velocity 2.0 s later is:

A stone thrown horizontally from a 4.9 m high wall lands after:

A student solves 4.9Δt² − 8.6Δt − 25 = 0 and gets Δt = 3.3 s or Δt = −1.5 s. The correct time of flight is:

For parents and teachers

Big idea: Projectile motion is two independent one-dimensional motions that share the same time (B2.2). Grade 12 adds launches from a height (a quadratic equation for time), velocities at impact and the derivation of the range equation. The activity is a predict-and-test inquiry (A1.5, A1.8).

How to help: Ask your teen to make a two-column table, horizontal and vertical, and list what is known in each before choosing an equation. The time is the only value that crosses between the columns.

Common difficulties:

Using the range equation when the landing height differs from the launch height.
Forgetting that the vertical velocity at the top is zero but the horizontal velocity is not.
Sign errors with Δy (a point below the launch is negative when up is positive).
Giving an impact speed as only the vertical component.

Safety: marbles are a choking hazard for small children.