IB Physics A.1 Kinematics | Lesson 4.2 | SL/HL

Angled Launch, Impact and Model Limits

Central question: How does the same component routine predict the whole flight—and where does the ideal model fail?

Learning Objectives and Success Criteria

Resolve

Signed launch components.

Reconstruct

Magnitude, direction and quadrant.

Evaluate

Qualitative resistance effects.

Retrieval | The L04-1 Component Routine

Use $g = 9.81 m s^{- 2}$ . Take right and upward as positive. Neglect fluid resistance unless a part asks about model limits.

(a) State $a_{x}$ and $a_{y}$ for ideal projectile motion.
(b) State the quantity shared by the two component models.
(c) Explain why the velocity vector is tangent to the trajectory.
(d) At the highest point, distinguish $v_{y}$ , total speed, and acceleration.

Reveal Q1 (1)

$a_{x} = 0$ , $a_{y} = - g$ .

Write the acceleration vector by components.

Reveal Q1 (2)

Time.

Link both models event by event.

Reveal Q1 (3)

Instantaneous velocity points in the direction of instantaneous displacement, which is tangent to the path.

Use the definition of instantaneous velocity.

Reveal Q1 (4)

$v_{y} = 0$ , total speed is $| v_{x} |$ and non-zero, and acceleration is $- g$ downward.

Do not confuse one zero component with a zero vector.

What Remains Unchanged?

Horizontal

a_{x} = 0

Vertical

a_{y} = - g

Connection

One shared time

Resolve the Launch Vector

Use $g = 9.81 m s^{- 2}$ . Take right and upward as positive. Neglect fluid resistance unless a part asks about model limits.

(a) A ball is launched at $20.0 m s^{- 1}$ , $30.0$ ° above horizontal. Calculate $u_{x}$ .
(b) Calculate $u_{y}$ .
(c) Explain how the sign of $u_{y}$ changes for a below-horizontal launch.

Reveal Q1 (1)

$17.3 m s^{- 1}$ .

Multiply the launch speed by the cosine of the stated angle.

Reveal Q1 (2)

$10.0 m s^{- 1}$ .

Multiply the launch speed by the sine of the stated angle.

Reveal Q1 (3)

It is negative with an upward-positive axis.

Resolve using the signed launch angle.

Above- and Below-Horizontal Signs

Above

u_{y} > 0

Below

u_{y} < 0

Keep upward positive for both cases; change the component sign, not the coordinate system.

Velocity Is Tangent to the Trajectory

The instantaneous velocity points along the direction of motion. Build that tangent vector from its horizontal and vertical components.

Track the Components Along the Path

Ascent

Positive vertical component decreases.

Top

Vertical component is zero.

Descent

Vertical component is negative.

Highest Point | One Component Is Zero

Use $g = 9.81 m s^{- 2}$ . Take right and upward as positive. Neglect fluid resistance unless a part asks about model limits.

(a) State the value of $v_{y}$ at the highest point.
(b) Explain why acceleration is still downward at that point.
(c) Calculate the time for the ball in p5 to reach its highest point.

Reveal Q2 (1)

$0 m s^{- 1}$ .

The vertical direction reverses here.

Reveal Q2 (2)

Gravity continues to act; the vertical velocity is only instantaneously zero.

Gravity does not depend on the instantaneous vertical velocity.

Reveal Q2 (3)

$1.02 s$ .

Use $0 = u_{y} - g t$ .

Acceleration Remains Downward

a_{y} = - g

Gravity does not switch off when the vertical velocity is instantaneously zero.

Build the Component-Knowns Table

Horizontal

u_x = u cos θ

Vertical

u_y = u sin θ

Guided Above-Horizontal Problem

Use $g = 9.81 m s^{- 2}$ . Take right and upward as positive. Neglect fluid resistance unless a part asks about model limits.

(a) Calculate its vertical displacement to the highest point.
(b) Calculate its horizontal displacement in that time.
(c) At $t = 1.50 s$ , calculate $v_{x}$ .

Reveal Q3 (1)

$5.10 m$ .

Use $v_{y}^{2} = u_{y}^{2} + 2 a_{y} Δ y$ .

Reveal Q3 (2)

$17.7 m$ .

Use $x = u_{x} t$ .

Reveal Q3 (3)

$17.3 m s^{- 1}$ .

With $a_{x} = 0$ , $v_{x} = u_{x}$ .

Time to Highest Point

0 = u_{y} - g t

Use the vertical component because the defining event is vertical.

Maximum Vertical Displacement

v_{y}^{2} = u_{y}^{2} + 2 a_{y} Δ y

Derive from the component knowns; no extra formula needs memorising.

Connect the Same Time Horizontally

Δ x = u_{x} t

The event time comes from the vertical story and is transferred unchanged.

Reconstruct Impact Velocity

Use $g = 9.81 m s^{- 2}$ . Take right and upward as positive. Neglect fluid resistance unless a part asks about model limits.

(a) At $t = 1.50 s$ , calculate $v_{y}$ .
(b) Explain how these components determine the quadrant of the velocity.
(c) A projectile has $v_{x} = 14.0 m s^{- 1}$ and $v_{y} = - 9.0 m s^{- 1}$ at impact. Calculate its speed.

Reveal Q4 (1)

$- 4.72 m s^{- 1}$ .

Use $v_{y} = u_{y} - g t$ .

Reveal Q4 (2)

Positive $v_{x}$ and negative $v_{y}$ place the velocity below the positive horizontal.

Inspect the signs before calculating an angle.

Reveal Q4 (3)

$16.6 m s^{- 1}$ .

Use $v = \sqrt{v_{x}^{2} + v_{y}^{2}}$ .

Check Magnitude, Angle and Quadrant

Use $g = 9.81 m s^{- 2}$ . Take right and upward as positive. Neglect fluid resistance unless a part asks about model limits.

(a) Calculate its direction below the horizontal.
(b) Explain why reporting only the speed is an incomplete impact velocity.
(c) Sketch $v_{x}$ against $t$ for the ideal flight.

Reveal Q5 (1)

$32.7$ ° below horizontal.

Use the inverse tangent of the component-magnitude ratio, then state the quadrant.

Reveal Q5 (2)

Velocity requires both magnitude and direction.

Give an angle and reference direction.

Reveal Q5 (3)

A constant positive horizontal line.

Its gradient is zero.

Four Component Graphs for the Trajectory

Horizontal pair

Linear position; constant velocity.

Vertical pair

Curved position; linearly decreasing velocity.

Below-Horizontal Launch | Same Method, Different Sign

u_{y} < 0

from the first instant; the vertical speed then grows downward under gravity.

Ideal Trajectory and Assumptions

The ideal path is parabolic when gravity is uniform and resistance is negligible. This statement is a model claim, not a guarantee for every real projectile.

Qualitative Effects of Fluid Resistance

Use $g = 9.81 m s^{- 2}$ . Take right and upward as positive. Neglect fluid resistance unless a part asks about model limits.

(a) Sketch $v_{y}$ against $t$ , including the highest point.
(b) Explain how the two graphs identify the instant when the projectile is moving horizontally.
(c) Explain two qualitative changes expected when fluid resistance is appreciable.

Reveal Q6 (1)

A straight line of gradient $- g$ crossing zero at the highest point.

Its gradient is constant and negative.

Reveal Q6 (2)

It occurs when the $v_{y}$ graph crosses zero while $v_{x}$ remains non-zero.

Align the time axes.

Reveal Q6 (3)

Horizontal speed decreases and the path is no longer a symmetric ideal parabola; range and maximum height are reduced for the same launch.

Drag opposes velocity, so it changes both components and destroys ideal symmetry.

Exit | What A.1 Predicts and A.2 Must Explain

Use $g = 9.81 m s^{- 2}$ . Take right and upward as positive. Neglect fluid resistance unless a part asks about model limits.

(a) Explain why a quantitative drag-force law is outside this lesson.
(b) State one prediction A.1 can make and one later A.2 question about its cause.

Reveal Q1 (1)

This lesson compares model predictions qualitatively; force-dependent acceleration belongs to A.2 dynamics.

No drag formula is needed here.

Reveal Q1 (2)

Example: A.1 predicts position or velocity; A.2 asks which resultant force produces the acceleration.

Separate kinematic description from dynamic cause.

Opening

Resolve and Track

Calculate and Reconstruct

Graphs and Model Limits

Angled Launch, Impact and Model Limits

Learning Objectives and Success Criteria

Resolve

Reconstruct

Evaluate

Retrieval | The L04-1 Component Routine

What Remains Unchanged?

Horizontal

Vertical

Connection

Resolve the Launch Vector

Above- and Below-Horizontal Signs

Above

Below

Velocity Is Tangent to the Trajectory

Track the Components Along the Path

Ascent

Top

Descent

Highest Point | One Component Is Zero

Acceleration Remains Downward

Build the Component-Knowns Table

Horizontal

Vertical

Guided Above-Horizontal Problem

Time to Highest Point

Maximum Vertical Displacement

Connect the Same Time Horizontally

Reconstruct Impact Velocity

Check Magnitude, Angle and Quadrant

Four Component Graphs for the Trajectory

Horizontal pair

Vertical pair

Below-Horizontal Launch | Same Method, Different Sign

Ideal Trajectory and Assumptions

Qualitative Effects of Fluid Resistance

Exit | What A.1 Predicts and A.2 Must Explain