Motion in two dimensions | AP Physics 1 Kinematics

In this sub-chapter, we study motion in two dimensions. We’ll focus on:

Horizontal and vertical equations of motion
Trajectory formulas
Relative motion

Decomposition of motion

A projectile launched with initial speed $v_{0}$ at an angle $θ$ above the horizontal has the velocity components:

$v_{0 x} = v_{0} cos θ, v_{0 y} = v_{0} sin θ .$

The key idea is that the motion splits into two independent one-dimensional motions:

Horizontal: No acceleration ( $a_{x} = 0$ ), so the horizontal velocity stays constant. The horizontal position is:

$x = v_{0 x} t .$

Vertical: Constant acceleration due to gravity ( $a_{y} = - g$ ), so the vertical position is:

$y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2} .$

Derivation of trajectory formulas

Here are the standard projectile-motion results, derived directly from the horizontal and vertical kinematics equations.

Time of flight (for launch from ground level)

Assume the projectile is launched from ground level ( $y_{0} = 0$ ) and lands back at ground level ( $y = 0$ ). The vertical displacement equation becomes:

$0 = v_{0 y} t - \frac{1}{2} g t^{2} .$

Factor out $t$ :

$t (v_{0 y} - \frac{1}{2} g t) = 0.$

The solution $t = 0$ corresponds to the launch instant, so we use the non-trivial solution:

$v_{0 y} - \frac{1}{2} g t = 0 ⟹ t = \frac{2 v _{0 y}}{g} = \frac{2 v _{0} sin θ}{g} .$

Thus, the time of flight is:

$T = \frac{2 v _{0} sin θ}{g} .$

Maximum height

At maximum height, the vertical velocity is zero. Using $v_{y} = v_{0 y} - g t$ :

$0 = v_{0 y} - g t_{up} .$

So the time to reach the top is:

$t_{up} = \frac{v _{0 y}}{g} = \frac{v _{0} sin θ}{g} .$

Now substitute $t_{up}$ into the vertical displacement equation (with $y_{0} = 0$ ):

$H = v_{0 y} t_{up} - \frac{1}{2} g t_{up}^{2} = v_{0} sin θ (\frac{v _{0} sin θ}{g}) - \frac{1}{2} g (\frac{v _{0} sin θ}{g})^{2} .$

Simplifying gives:

$H = \frac{v _{0}^{2} sin ^{2} θ}{g} - \frac{1}{2} \frac{v _{0}^{2} sin ^{2} θ}{g} = \frac{v _{0}^{2} sin ^{2} θ}{2 g} .$

Horizontal range

The horizontal range is the horizontal displacement after time $T$ :

$R = v_{0 x} T = v_{0} cos θ (\frac{2 v _{0} sin θ}{g}) = \frac{2 v _{0}^{2} sin θ cos θ}{g} .$

Using $sin 2 θ = 2 sin θ cos θ$ :

$R = \frac{v _{0}^{2} sin 2 θ}{g} .$

In the practice problems below, the goal is to rely on the basic horizontal and vertical motion equations and use them to derive what you need, rather than depending only on memorized formulas. Example Problem 1

A projectile is launched from the ground with an initial speed of $15 m/s$ at an angle of $4 0^{\circ}$ . Determine:

The time of flight.
The maximum height reached.
The horizontal range.

Solution:

(spoiler)

Time of flight: Start with the vertical displacement equation:

$y = v_{0 y} t - \frac{1}{2} g t^{2} .$

Since the projectile starts and lands at ground level, $y_{0} = 0$ and the final $y = 0$ , so:

$0 = (15 sin 4 0^{\circ}) t - \frac{1}{2} (9.8) t^{2} .$

Factor out $t$ and ignore the trivial solution $t = 0$ :

$(15 sin 4 0^{\circ}) - \frac{1}{2} (9.8) t = 0.$

Solve for $t$ :

$t = \frac{2 ( 15 sin 4 0 ^{\circ} )}{9.8} .$

Using $sin 4 0^{\circ} \approx 0.643$ :

$T \approx \frac{2 ( 15 \times 0.643 )}{9.8} \approx \frac{19.29}{9.8} \approx 1.97 s .$
Maximum height: At the top of the trajectory, $v_{y} = 0$ . Using:

$v_{y} = v_{0 y} - g t,$

we get:

$t_{up} = \frac{15 sin 4 0 ^{\circ}}{9.8} \approx \frac{15 \times 0.643}{9.8} \approx \frac{9.645}{9.8} \approx 0.98 s .$

Substitute into the vertical displacement equation:

$H = (15 sin 4 0^{\circ}) (0.98) - \frac{1}{2} (9.8) (0.98)^{2} .$

Evaluate:

$H \approx 9.645 \times 0.98 - 4.9 \times 0.96 \approx 9.45 - 4.70 \approx 4.75 m .$
Horizontal range: Horizontal motion has constant velocity, so:

$x = v_{0 x} t,$

where $v_{0 x} = 15 cos 4 0^{\circ}$ . Using $cos 4 0^{\circ} \approx 0.766$ :

$v_{0 x} \approx 15 \times 0.766 \approx 11.49 m/s .$

Then:

$R = 11.49 \times 1.97 \approx 22.62 m .$

Example problem 2

A ball is thrown from the top of a $5 m$ high platform with an initial speed of $20 m/s$ at an angle of $3 0^{\circ}$ above the horizontal. Determine:

The time until the ball hits the ground.
The horizontal distance from the platform where the ball lands.

Solution:

(spoiler)

Graphing the given two dimensional motion scenario

Decompose the initial velocity:

$v_{0 x} = 20 cos 3 0^{\circ} \approx 20 (0.866) = 17.32 m/s,$

$v_{0 y} = 20 sin 3 0^{\circ} = 20 (0.5) = 10 m/s .$
Vertical motion: Use:

$y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2},$

with $y_{0} = 5 m$ and final $y = 0$ :

$0 = 5 + 10 t - 4.9 t^{2} .$

Rearrange into standard quadratic form:

$4.9 t^{2} - 10 t - 5 = 0.$

Apply the quadratic formula:

$t = \frac{10 \pm 1 0 ^{2} - 4 ( 4.9 ) ( - 5 )}{2 ( 4.9 )} .$

Simplify:

$t = \frac{10 \pm 100 + 98}{9.8} = \frac{10 \pm 198}{9.8} .$

Take the positive root (time must be positive):

$t \approx \frac{10 + 14.07}{9.8} \approx \frac{24.07}{9.8} \approx 2.46 s .$
Horizontal motion: Now use $x = v_{0 x} t$ :

$x = v_{0 x} t \approx 17.32 m/s \times 2.46 s \approx 42.59 m .$

Example problem 3

A projectile is launched from the ground with an initial speed of $18 m/s$ at an angle of $6 0^{\circ}$ . When the projectile is at half its maximum height, determine the angle of its velocity vector with respect to the horizontal.

Solution:

(spoiler)

Determine maximum height: Start by decomposing the initial velocity:

$v_{0 x} = 18 cos 6 0^{\circ} = 18 (0.5) = 9 m/s, v_{0 y} = 18 sin 6 0^{\circ} \approx 18 (0.866) = 15.59 m/s .$

Then compute the maximum height:

$H = \frac{( v _{0 y} ) ^{2}}{2 g} = \frac{( 15.59 ) ^{2}}{19.6} \approx \frac{243.1}{19.6} \approx 12.41 m .$

Half of the maximum height is about $6.205 m$ .
Find the time at half-height: Use:

$y = v_{0 y} t - \frac{1}{2} g t^{2} .$

Set $y = 6.205 m$ :

$6.205 = 15.59 t - 4.9 t^{2} .$

Solve this quadratic for $t$ and use the smaller positive solution (the projectile is on the way up), say $t \approx 0.8 s$ .
Determine velocity components at $t \approx 0.8 s$ : Horizontal velocity stays constant:

$v_{x} = v_{0 x} = 9 m/s .$

Vertical velocity is:

$v_{y} = v_{0 y} - g t \approx 15.59 - 9.8 (0.8) \approx 15.59 - 7.84 \approx 7.75 m/s .$
Compute the angle: Use the velocity components:

$tan ϕ = \frac{v _{y}}{v _{x}} \approx \frac{7.75}{9} \approx 0.8611,$

$ϕ \approx arctan (0.8611) \approx 4 0^{\circ} .$

Relative motion

Relative motion describes how an object’s velocity depends on the observer’s frame of reference.

If object $A$ has velocity $v_{A}$ relative to the ground, and object $B$ has velocity $v_{B}$ relative to the ground, then the velocity of $A$ relative to $B$ is:

$v_{A / B} = v_{A} - v_{B} .$

A common example is a swimmer crossing a river. The swimmer’s velocity relative to the ground is the vector sum of:

$v_{s}$ , the swimmer’s velocity relative to the water
$v_{r}$ , the river current (water relative to the ground)

So:

$v_{s / g} = v_{s} + v_{r} .$

This is the main idea to keep in mind: the swimmer’s path relative to the ground depends on both the swimmer’s motion through the water and the water’s motion relative to the ground.

Vector representation of swimmer's velocity and river flow

Case 1: Swimmer aiming to reach the opposite bank

To land directly opposite the starting point, the swimmer must aim upstream so that the upstream component of the swimming velocity cancels the river current.

Let:

$v_{s}$ be the swimmer’s speed relative to the water.
$v_{r}$ be the speed of the river.
$θ$ be the angle upstream (measured from the line perpendicular to the bank).

For no downstream drift, the upstream component must match the river current:

$v_{s} sin θ = v_{r} ⟹ θ = arcsin (\frac{v _{r}}{v _{s}}) .$

The effective speed across the river is the component perpendicular to the bank:

$v_{s, ⊥} = v_{s} cos θ .$

For a river of width $w$ , the crossing time is:

$t = \frac{w}{v _{s, ⊥}} = \frac{w}{v _{s} cos θ} .$

Case 2: Swimmer swimming straight across

If the swimmer points straight across (perpendicular to the banks), then the swimmer’s velocity relative to the water is entirely across the river:

$v_{s, ⊥} = v_{s} .$

The river current still carries the swimmer downstream. The downstream displacement is:

$d = v_{r} t,$

where the crossing time is:

$t = \frac{w}{v _{s}} .$

Example problem 4

A swimmer can swim at $2.0 m/s$ in still water and needs to cross a river that is $10 m$ wide. The river current is $0.8 m/s$ . At what angle (measured from the perpendicular to the bank) must the swimmer aim to land directly opposite the starting point? Also, determine the time to cross the river.

Solution:

(spoiler)

To cancel the river current:

$v_{s} sin θ = 0.8 ⟹ sin θ = \frac{0.8}{2.0} = 0.4,$

$θ = arcsin (0.4) \approx 23. 6^{\circ} .$

The effective speed across the river is:

$v_{s, ⊥} = v_{s} cos θ \approx 2.0 cos (23. 6^{\circ}) \approx 2.0 \times 0.917 \approx 1.83 m/s .$

Time to cross:

$t = \frac{10 m}{1.83 m/s} \approx 5.46 s .$

Example problem 5

Using the same swimmer (swimming at $2.0 m/s$ ) and a river current of $0.8 m/s$ , if the swimmer heads straight across (perpendicular to the banks), how far downstream will he be carried?

Solution:

(spoiler)

Time to cross:

$t = \frac{10 m}{2.0 m/s} = 5.0 s .$

Downstream drift:

$d = v_{r} t = 0.8 m/s \times 5.0 s = 4.0 m .$

In this sub-chapter, we study motion in two dimensions. We’ll focus on:

Horizontal and vertical equations of motion
Trajectory formulas
Relative motion

Decomposition of motion

A projectile launched with initial speed $v_{0}$ at an angle $θ$ above the horizontal has the velocity components:

$v_{0 x} = v_{0} cos θ, v_{0 y} = v_{0} sin θ .$

The key idea is that the motion splits into two independent one-dimensional motions:

Horizontal: No acceleration ( $a_{x} = 0$ ), so the horizontal velocity stays constant. The horizontal position is:

$x = v_{0 x} t .$

Vertical: Constant acceleration due to gravity ( $a_{y} = - g$ ), so the vertical position is:

$y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2} .$

Derivation of trajectory formulas

Here are the standard projectile-motion results, derived directly from the horizontal and vertical kinematics equations.

Time of flight (for launch from ground level)

Assume the projectile is launched from ground level ( $y_{0} = 0$ ) and lands back at ground level ( $y = 0$ ). The vertical displacement equation becomes:

$0 = v_{0 y} t - \frac{1}{2} g t^{2} .$

Factor out $t$ :

$t (v_{0 y} - \frac{1}{2} g t) = 0.$

The solution $t = 0$ corresponds to the launch instant, so we use the non-trivial solution:

$v_{0 y} - \frac{1}{2} g t = 0 ⟹ t = \frac{2 v _{0 y}}{g} = \frac{2 v _{0} sin θ}{g} .$

Thus, the time of flight is:

$T = \frac{2 v _{0} sin θ}{g} .$

Maximum height

At maximum height, the vertical velocity is zero. Using $v_{y} = v_{0 y} - g t$ :

$0 = v_{0 y} - g t_{up} .$

So the time to reach the top is:

$t_{up} = \frac{v _{0 y}}{g} = \frac{v _{0} sin θ}{g} .$

Now substitute $t_{up}$ into the vertical displacement equation (with $y_{0} = 0$ ):

$H = v_{0 y} t_{up} - \frac{1}{2} g t_{up}^{2} = v_{0} sin θ (\frac{v _{0} sin θ}{g}) - \frac{1}{2} g (\frac{v _{0} sin θ}{g})^{2} .$

Simplifying gives:

$H = \frac{v _{0}^{2} sin ^{2} θ}{g} - \frac{1}{2} \frac{v _{0}^{2} sin ^{2} θ}{g} = \frac{v _{0}^{2} sin ^{2} θ}{2 g} .$

Horizontal range

The horizontal range is the horizontal displacement after time $T$ :

$R = v_{0 x} T = v_{0} cos θ (\frac{2 v _{0} sin θ}{g}) = \frac{2 v _{0}^{2} sin θ cos θ}{g} .$

Using $sin 2 θ = 2 sin θ cos θ$ :

$R = \frac{v _{0}^{2} sin 2 θ}{g} .$

A projectile is launched from the ground with an initial speed of $15 m/s$ at an angle of $4 0^{\circ}$ . Determine:

The time of flight.
The maximum height reached.
The horizontal range.

Solution:

(spoiler)

Time of flight: Start with the vertical displacement equation:

$y = v_{0 y} t - \frac{1}{2} g t^{2} .$

Since the projectile starts and lands at ground level, $y_{0} = 0$ and the final $y = 0$ , so:

$0 = (15 sin 4 0^{\circ}) t - \frac{1}{2} (9.8) t^{2} .$

Factor out $t$ and ignore the trivial solution $t = 0$ :

$(15 sin 4 0^{\circ}) - \frac{1}{2} (9.8) t = 0.$

Solve for $t$ :

$t = \frac{2 ( 15 sin 4 0 ^{\circ} )}{9.8} .$

Using $sin 4 0^{\circ} \approx 0.643$ :

$T \approx \frac{2 ( 15 \times 0.643 )}{9.8} \approx \frac{19.29}{9.8} \approx 1.97 s .$
Maximum height: At the top of the trajectory, $v_{y} = 0$ . Using:

$v_{y} = v_{0 y} - g t,$

we get:

$t_{up} = \frac{15 sin 4 0 ^{\circ}}{9.8} \approx \frac{15 \times 0.643}{9.8} \approx \frac{9.645}{9.8} \approx 0.98 s .$

Substitute into the vertical displacement equation:

$H = (15 sin 4 0^{\circ}) (0.98) - \frac{1}{2} (9.8) (0.98)^{2} .$

Evaluate:

$H \approx 9.645 \times 0.98 - 4.9 \times 0.96 \approx 9.45 - 4.70 \approx 4.75 m .$
Horizontal range: Horizontal motion has constant velocity, so:

$x = v_{0 x} t,$

where $v_{0 x} = 15 cos 4 0^{\circ}$ . Using $cos 4 0^{\circ} \approx 0.766$ :

$v_{0 x} \approx 15 \times 0.766 \approx 11.49 m/s .$

Then:

$R = 11.49 \times 1.97 \approx 22.62 m .$

Example problem 2

A ball is thrown from the top of a $5 m$ high platform with an initial speed of $20 m/s$ at an angle of $3 0^{\circ}$ above the horizontal. Determine:

The time until the ball hits the ground.
The horizontal distance from the platform where the ball lands.

Solution:

(spoiler)

Decompose the initial velocity:

$v_{0 x} = 20 cos 3 0^{\circ} \approx 20 (0.866) = 17.32 m/s,$

$v_{0 y} = 20 sin 3 0^{\circ} = 20 (0.5) = 10 m/s .$
Vertical motion: Use:

$y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2},$

with $y_{0} = 5 m$ and final $y = 0$ :

$0 = 5 + 10 t - 4.9 t^{2} .$

Rearrange into standard quadratic form:

$4.9 t^{2} - 10 t - 5 = 0.$

Apply the quadratic formula:

$t = \frac{10 \pm 1 0 ^{2} - 4 ( 4.9 ) ( - 5 )}{2 ( 4.9 )} .$

Simplify:

$t = \frac{10 \pm 100 + 98}{9.8} = \frac{10 \pm 198}{9.8} .$

Take the positive root (time must be positive):

$t \approx \frac{10 + 14.07}{9.8} \approx \frac{24.07}{9.8} \approx 2.46 s .$
Horizontal motion: Now use $x = v_{0 x} t$ :

$x = v_{0 x} t \approx 17.32 m/s \times 2.46 s \approx 42.59 m .$

Example problem 3

Solution:

(spoiler)

Determine maximum height: Start by decomposing the initial velocity:

$v_{0 x} = 18 cos 6 0^{\circ} = 18 (0.5) = 9 m/s, v_{0 y} = 18 sin 6 0^{\circ} \approx 18 (0.866) = 15.59 m/s .$

Then compute the maximum height:

$H = \frac{( v _{0 y} ) ^{2}}{2 g} = \frac{( 15.59 ) ^{2}}{19.6} \approx \frac{243.1}{19.6} \approx 12.41 m .$

Half of the maximum height is about $6.205 m$ .
Find the time at half-height: Use:

$y = v_{0 y} t - \frac{1}{2} g t^{2} .$

Set $y = 6.205 m$ :

$6.205 = 15.59 t - 4.9 t^{2} .$

Solve this quadratic for $t$ and use the smaller positive solution (the projectile is on the way up), say $t \approx 0.8 s$ .
Determine velocity components at $t \approx 0.8 s$ : Horizontal velocity stays constant:

$v_{x} = v_{0 x} = 9 m/s .$

Vertical velocity is:

$v_{y} = v_{0 y} - g t \approx 15.59 - 9.8 (0.8) \approx 15.59 - 7.84 \approx 7.75 m/s .$
Compute the angle: Use the velocity components:

$tan ϕ = \frac{v _{y}}{v _{x}} \approx \frac{7.75}{9} \approx 0.8611,$

$ϕ \approx arctan (0.8611) \approx 4 0^{\circ} .$

Relative motion

Relative motion describes how an object’s velocity depends on the observer’s frame of reference.

If object $A$ has velocity $v_{A}$ relative to the ground, and object $B$ has velocity $v_{B}$ relative to the ground, then the velocity of $A$ relative to $B$ is:

$v_{A / B} = v_{A} - v_{B} .$

A common example is a swimmer crossing a river. The swimmer’s velocity relative to the ground is the vector sum of:

$v_{s}$ , the swimmer’s velocity relative to the water
$v_{r}$ , the river current (water relative to the ground)

So:

$v_{s / g} = v_{s} + v_{r} .$

This is the main idea to keep in mind: the swimmer’s path relative to the ground depends on both the swimmer’s motion through the water and the water’s motion relative to the ground.

Case 1: Swimmer aiming to reach the opposite bank

To land directly opposite the starting point, the swimmer must aim upstream so that the upstream component of the swimming velocity cancels the river current.

Let:

$v_{s}$ be the swimmer’s speed relative to the water.
$v_{r}$ be the speed of the river.
$θ$ be the angle upstream (measured from the line perpendicular to the bank).

For no downstream drift, the upstream component must match the river current:

$v_{s} sin θ = v_{r} ⟹ θ = arcsin (\frac{v _{r}}{v _{s}}) .$

The effective speed across the river is the component perpendicular to the bank:

$v_{s, ⊥} = v_{s} cos θ .$

For a river of width $w$ , the crossing time is:

$t = \frac{w}{v _{s, ⊥}} = \frac{w}{v _{s} cos θ} .$

Case 2: Swimmer swimming straight across

If the swimmer points straight across (perpendicular to the banks), then the swimmer’s velocity relative to the water is entirely across the river:

$v_{s, ⊥} = v_{s} .$

The river current still carries the swimmer downstream. The downstream displacement is:

$d = v_{r} t,$

where the crossing time is:

$t = \frac{w}{v _{s}} .$

Example problem 4

Solution:

(spoiler)

To cancel the river current:

$v_{s} sin θ = 0.8 ⟹ sin θ = \frac{0.8}{2.0} = 0.4,$

$θ = arcsin (0.4) \approx 23. 6^{\circ} .$

The effective speed across the river is:

$v_{s, ⊥} = v_{s} cos θ \approx 2.0 cos (23. 6^{\circ}) \approx 2.0 \times 0.917 \approx 1.83 m/s .$

Time to cross:

$t = \frac{10 m}{1.83 m/s} \approx 5.46 s .$

Example problem 5

Using the same swimmer (swimming at $2.0 m/s$ ) and a river current of $0.8 m/s$ , if the swimmer heads straight across (perpendicular to the banks), how far downstream will he be carried?

Solution:

(spoiler)

Time to cross:

$t = \frac{10 m}{2.0 m/s} = 5.0 s .$

Downstream drift:

$d = v_{r} t = 0.8 m/s \times 5.0 s = 4.0 m .$