Why is 45 degrees the angle for maximum range on level ground?

Because R equals u squared sin 2 theta divided by g, and sin 2 theta reaches its maximum of 1 when 2 theta equals 90 degrees, i.e., theta equals 45 degrees. Maximum range is u squared divided by g.

At the highest point of a projectile trajectory, what are its velocity and acceleration?

Vertical velocity is zero but horizontal velocity u cos theta is unchanged. So speed at the apex is u cos theta, not zero. Acceleration is g downward everywhere on the trajectory.

Motion In Two And Three Dimension

Q: Why is the path of a projectile a parabola?

Because horizontal displacement grows linearly with time while vertical displacement is quadratic. Eliminating t gives y as a quadratic function of x, which is a parabola.

Q: Do two projectiles fired at complementary angles have equal ranges?

Yes on a flat horizontal plane. Angles theta and 90 minus theta give the same range because sin 2(90-theta) equals sin 2 theta. But their times of flight and maximum heights differ.

Q: How does air resistance change projectile motion?

It reduces range and maximum height, shortens time of flight, and makes the trajectory asymmetric with a steeper descending arm. Standard textbook problems neglect it.

Q: What is the difference between the position vector and displacement vector?

The position vector points from the origin to the current location and depends on the choice of origin. Displacement is the change in position between two instants and does not depend on the origin.

Q: Can projectile motion be treated as two independent 1-D motions?

Yes. Horizontal motion has zero acceleration and constant velocity u cos theta; vertical motion has acceleration minus g and initial vertical velocity u sin theta. Solve each separately and combine.

PhysicsKinematicsFor JEE aspirants

Motion in two and three dimensions is the motion of a particle in a plane (2-D) or in space (3-D), where position, velocity, and acceleration are described by two or three components respectively. This concept covers the vector formulation of kinematics, position vector and displacement, average and instantaneous velocity in a plane, projectile motion (time of flight, range, maximum height, trajectory equation), and projectile motion on an inclined plane - core topics for JEE Physics and NEET Physics.

Key Formulas - Quick Reference

Position vector (2-D): $r = x \hat{i} + y \hat{j}$
Displacement: $Δ r = r_{2} - r_{1}$
Instantaneous velocity: $v = \frac{d r}{d t} = v_{x} \hat{i} + v_{y} \hat{j}$
Projectile - Time of flight: $T = \frac{2 u sin θ}{g}$
Projectile - Horizontal range: $R = \frac{u ^{2} sin 2 θ}{g}$ , $R_{ma x} = \frac{u ^{2}}{g}$ at $θ = 45°$
Projectile - Maximum height: $H = \frac{u ^{2} sin ^{2} θ}{2 g}$
Trajectory equation: $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} cos ^{2} θ}$ (parabola)
On incline - Time of flight: $T = \frac{2 u sin θ}{g cos α}$
On incline - Max range: $R_{ma x} = \frac{u ^{2}}{g ( 1 \pm sin α )}$ (+ for down-plane, − for up-plane)

1. Motion in Two and Three Dimensions

Motion of a particle in a plane is called two-dimensional (2-D) motion. Its position, velocity, and acceleration can have two non-zero components in any two mutually perpendicular directions of a Cartesian coordinate system. Motion in space is three-dimensional (3-D), needing three components. In either case, vectors are the natural tool.

1.1 Vectors and scalars - quick recap

A vector has both magnitude and direction (e.g., velocity, acceleration, displacement, force).
A scalar has magnitude only (e.g., mass, temperature, speed, distance).
A zero vector has magnitude zero and no fixed direction.

1.2 Scalar and vector products

Scalar (dot) product:

A \cdot B = ∣ A ∣ ∣ B ∣ cos θ

Vector (cross) product:

∣ A \times B ∣ = ∣ A ∣ ∣ B ∣ sin θ

In component form:

A = a_{x} \hat{i} + a_{y} \hat{j} + a_{z} \hat{k}

, magnitude

∣ A ∣ = a_{x}^{2} + a_{y}^{2} + a_{z}^{2}

2. Position and Displacement Vectors

Figure 6: Position vector r⃗ = xî + yĵ of a particle in a 2D plane.

The position of a particle P in a 2-D plane at any instant is:

r = x \hat{i} + y \hat{j}

Here $x \hat{i}$ and $y \hat{j}$ are the vector components of $r$ along the x and y axes, and the scalars $x, y$ give the location of the particle. For example, $r = 3 \hat{i} + 6 \hat{j}$ locates P at $(3, 6)$ .

If the particle has position vectors $r_{1}$ at time $t_{1}$ and $r_{2}$ at time $t_{2}$ , its displacement is:

Δ r = r_{2} - r_{1}

3. Velocity in Two Dimensions

If the particle undergoes a displacement $Δ r$ in time $Δ t$ , the average velocity is:

⟨ v ⟩ = \frac{Δ r}{Δ t} = \frac{Δ x}{Δ t} \hat{i} + \frac{Δ y}{Δ t} \hat{j}

The instantaneous velocity is the limit as $Δ t \to 0$ :

v = Δ t \to 0 lim \frac{Δ r}{Δ t} = \frac{d r}{d t} = \frac{d x}{d t} \hat{i} + \frac{d y}{d t} \hat{j} = v_{x} \hat{i} + v_{y} \hat{j}

Note on direction: As

Δ t \to 0

, the direction of

Δ r

approaches the tangent to the path. So instantaneous velocity is always tangent to the particle's trajectory. In 1-D motion the sign of the slope of the x-t graph gave both magnitude and direction. In 2-D motion, the x-y graph (trajectory) shows only the direction (tangent), not the speed.

Solved Example 1

A particle of mass

1 0^{- 2}

kg moves along the positive x-axis under a force

F (x) = - \frac{k}{2 x ^{2}}

where

k = 1 0^{- 2} N m^{2}

. At

t = 0

, it is at

x = 1.0

m with velocity

v = 0

. Find its velocity when it reaches

x = 0.5

Solution:

Force per unit mass gives the acceleration:

a = \frac{F}{m} = - \frac{k}{2 m x ^{2}} = - \frac{1 0 ^{- 2}}{2 \times 1 0 ^{- 2} \times x ^{2}} = - \frac{1}{2 x ^{2}}

Use $v \frac{d v}{d x} = a$ , i.e. $v d v = - \frac{1}{2} \frac{d x}{x ^{2}}$ . Integrate from $(x = 1, v = 0)$ to $(x = 0.5, v = v)$ :

\int_{0}^{v} v d v = - \frac{1}{2} \int_{1.0}^{0.5} \frac{d x}{x ^{2}} \Rightarrow \frac{v ^{2}}{2} = \frac{1}{2} [\frac{1}{x}]_{1.0}^{0.5} = \frac{1}{2} (2 - 1) = \frac{1}{2}

So $v^{2} = 1.0 \Rightarrow ∣ v ∣ = 1.0$ m/s. Since motion is in the negative-x direction, $v = - 1.0 m/s$ .

Solved Example 2

A particle moves such that

a = A - B v

, where

A

and

B

are constants and

v

is the velocity. Find the velocity as a function of time and the terminal velocity.

Solution:

Write $\frac{d v}{d t} = A - B v$ and separate variables:

\int_{0}^{V} \frac{d v}{A - B v} = \int_{0}^{t} d t \Rightarrow - \frac{1}{B} lo g_{e} (\frac{A - B V}{A}) = t

\Rightarrow A - B V = A e^{- B t} \Rightarrow V (t) = \frac{A}{B} (1 - e^{- B t})

At terminal velocity, acceleration is zero: $A - B v_{t} = 0 \Rightarrow v_{t} = A / B$ .

Solved Example 3

A body slides from rest down a frictionless incline. Its velocity down the plane is maximum at which angle of inclination? (a)

0°

(b)

90°

(c)

360°

(d)

180°

Solution: (b)

Using $v = (g sin θ) t$ , $v$ is maximum when $sin θ = 1$ , i.e., $θ = 90°$ .

Solved Example 4

The radius vector of a point depends on time as

r = c t + \frac{b t ^{2}}{2}

, where

c

and

b

are constant vectors. Find the magnitudes of velocity and acceleration at any time

t

Solution:

Velocity: $v = \frac{d r}{d t} = c + b t$ .

Magnitude: using $c \cdot c = c^{2}$ , $b \cdot b = b^{2}$ , and $c \cdot b = c b cos θ$ :

∣ v ∣ = c^{2} + b^{2} t^{2} + 2 c b t cos θ

Acceleration: $a = \frac{d v}{d t} = b$ , so $∣ a ∣ = ∣ b ∣ = b$ .

Solved Example 5

The position vector of a particle is initially

r_{1} = - 2 \hat{i} - 2 \hat{j}

and later

r_{2} = 8 \hat{i} + 10 \hat{j}

. What is the displacement?

Solution:

Δ r = r_{2} - r_{1} = (8 \hat{i} + 10 \hat{j}) - (- 2 \hat{i} - 2 \hat{j}) = 10 \hat{i} + 12 \hat{j}

Magnitude: $∣Δ r ∣ = 1 0^{2} + 1 2^{2} = 244 \approx 15.62$ m.

4. Projectile Motion

If an object is given an initial velocity in any direction (except purely vertical) and then allowed to move freely under gravity alone, its motion is called projectile motion. In the standard analysis:

Acceleration due to gravity $g$ is constant and directed vertically downward.
Air resistance is neglected.
The Earth's surface is treated as flat.

Projectile motion is 2-D motion and can be treated as the superposition of two independent motions:

Horizontal: constant velocity $u cos θ$ , acceleration $= 0$ .
Vertical: initial velocity $u sin θ$ , acceleration $= - g$ .

Figure 7: Projectile trajectory - parabolic path with range R and maximum height H.

4.1 Main features of a projectile

(i) The particle moves horizontally and vertically at the same time - the resulting path is a curve.
(ii) Origin is usually taken at the launch point, with $x$ -axis horizontal and $y$ -axis vertical.
(iii) The velocity at any instant is tangent to the path with horizontal and vertical components.
(iv) The only force acting is weight $m g$ downward, so acceleration is $g$ vertically downward. Air friction is neglected.
(v) Because $g$ does not change with time, projectile motion is uniformly accelerated at every instant: $a_{x} = 0$ , $a_{y} = - g$ .

4.2 Time of flight, horizontal range, maximum height

Figure 8: Velocity components along the trajectory - horizontal stays constant, vertical reverses.

Let $P (x, y)$ be the projectile's position at time $t$ after projection. At the launch point O, horizontal velocity $= u cos θ$ (constant throughout) and vertical velocity $= u sin θ$ initially.

Time of flight $T$ : The time from launch until the projectile returns to the same horizontal level.

Set vertical displacement equal to zero:

0 = u sin θ t - \frac{1}{2} g t^{2} = t (u sin θ - \frac{1}{2} g t)

So $t = 0$ (launch) or:

T = \frac{2 u sin θ}{g}

Horizontal range $R$ : The horizontal distance covered during the time of flight.

R = u cos θ \times T = \frac{u ^{2} sin 2 θ}{g}

R_{ma x} = \frac{u ^{2}}{g} at θ = 45°

For a given launch speed $u$ , the same range is obtained for angles $θ$ and $(90° - θ)$ - since $sin 2 (90° - θ) = sin (180° - 2 θ) = sin 2 θ$ .

Maximum height $H$ : Achieved when vertical velocity component is zero.

Using $v_{y}^{2} = (u sin θ)^{2} - 2 g H$ with $v_{y} = 0$ :

H_{ma x} = \frac{u ^{2} sin ^{2} θ}{2 g}

4.3 Velocity of the projectile at time $t$

v = u cos θ \hat{i} + (u sin θ - g t) \hat{j}

Speed (magnitude):

∣ v ∣ = u^{2} + g^{2} t^{2} - 2 u g t sin θ

4.4 Trajectory equation

Eliminate $t$ from the horizontal and vertical position equations.

Vertical: $y = u sin θ t - \frac{1}{2} g t^{2}$ ; horizontal: $x = u cos θ t$ , so $t = x / (u cos θ)$ . Substituting:

y = x tan θ - \frac{g x ^{2}}{2 u ^{2} cos ^{2} θ}

This is the equation of a parabola - the trajectory of a projectile is always parabolic (under uniform gravity, ignoring air resistance).

Solved Example 6

Two positions A and B lie at the same height

h

above the ground on the trajectory of a projectile. The maximum height of the projectile is

H

. Find the time

t

elapsed between the positions A and B in terms of

H

and

h

Solution:

Let $T$ be the total time of flight. If A is at time $t_{1}$ and B at time $t_{2}$ (both at height $h$ ), symmetry gives $t_{1} + t_{2} = T$ and $t = t_{2} - t_{1}$ .

Solving $h = u sin θ t^{'} - \frac{1}{2} g t^{'2}$ for $t^{'}$ gives two roots whose sum is $T$ and product is $2 h / g$ . Then:

t^{2} = (t_{2} - t_{1})^{2} = (t_{1} + t_{2})^{2} - 4 t_{1} t_{2} = T^{2} - \frac{8 h}{g}

Since $T^{2} = \frac{4 u ^{2} sin ^{2} θ}{g ^{2}} = \frac{8 H}{g}$ (using $H = u^{2} sin^{2} θ / (2 g)$ ):

t = \frac{8}{g} (H - h)

5. Projectile Motion on an Inclined Plane

Figure 9: Projectile on an inclined plane - g resolved along and perpendicular to the incline.

Consider an inclined plane of inclination $α$ and a particle launched from a point on the plane with speed $u$ at angle $θ$ measured from the surface of the plane. Take the x-axis along the plane (up the slope) and the y-axis perpendicular to it.

Gravity is resolved into two perpendicular components:

Along the plane (down the slope): $g sin α$ - this decelerates $u_{x} = u cos θ$ (for up-slope motion) or accelerates it (for down-slope).
Perpendicular to the plane: $g cos α$ - decelerates $u_{y} = u sin θ$ .

5.1 Time of flight (return to inclined surface)

The perpendicular displacement is zero at launch and return. Using $0 = u sin θ T - \frac{1}{2} (g cos α) T^{2}$ :

T = \frac{2 u sin θ}{g cos α}

5.2 Maximum height above the inclined surface

H = \frac{u ^{2} sin ^{2} θ}{2 g cos α}

5.3 Range along the inclined plane

Because the x-axis also has an acceleration $\pm g sin α$ , we cannot use the simple range formula. Compute:

R = u cos θ \cdot T - \frac{1}{2} (g sin α) T^{2}

Substituting the value of $T$ :

R_{u p} = \frac{u ^{2} sin 2 θ}{g cos α} - \frac{2 u ^{2} sin α sin ^{2} θ}{g cos ^{2} α} (projectile up the plane)

R_{d o w n} = \frac{u ^{2} sin 2 θ}{g cos α} + \frac{2 u ^{2} sin α sin ^{2} θ}{g cos ^{2} α} (projectile down the plane)

Advice: Don't memorise the range expression as a standard result. Set up the two-axis analysis for each problem, since the sign of

g sin α

flips depending on the direction of motion up or down the slope.

5.4 Maximum range on the inclined plane

Setting $d R / d θ = 0$ gives:

R_{ma x} = \frac{u ^{2}}{g ( 1 \pm sin α )}

Here the $+$ sign is for projectile up the plane; the $-$ sign is for projectile down the plane.

Solved Example 7

A ball is dropped from a height

h

above a point on an inclined plane of inclination

θ

. The ball collides elastically with the surface and rebounds. Find the distance from the point of first impact to the point where the ball hits the plane a second time.

Solution:

Take the first impact point as origin. Let x-axis be along the plane (down-slope) and y-axis perpendicular to the plane. The ball hits the plane with speed $u = 2 g h$ . After an elastic collision, the perpendicular component of velocity reverses (magnitude $u cos θ$ ) and the along-plane component ( $u sin θ$ ) is unchanged. So after rebound:

$u_{x} = u sin θ$ (along-plane), $u_{y} = u cos θ$ (perpendicular).

Along-plane, the acceleration is $g sin θ$ (down-slope). Perpendicular, it is $- g cos θ$ .

Time from first impact to second impact (perpendicular displacement returns to 0):

T = \frac{2 u _{y}}{g cos θ} = \frac{2 u cos θ}{g cos θ} = \frac{2 u}{g}

Distance along the plane in this time:

R = u_{x} T + \frac{1}{2} (g sin θ) T^{2} = u sin θ (\frac{2 u}{g}) + \frac{1}{2} (g sin θ) (\frac{2 u}{g})^{2} = \frac{4 u ^{2} sin θ}{g}

With $u^{2} = 2 g h$ : $R = 8 h sin θ$ .

Solved Example 8

A projectile is thrown with speed

u

at angle

θ

to an inclined plane of inclination

β

. Find the angle

θ

at which the projectile must be thrown so that it strikes the inclined plane normally.

Solution:

Take axes along and perpendicular to the plane. Time of flight:

T = \frac{2 u sin θ}{g cos β}

Along-plane velocity as a function of time: $v_{x} (t) = u cos θ - g sin β \cdot t$ . Striking normally means the along-plane component is zero at impact ( $v_{x} (T) = 0$ ):

0 = u cos θ - g sin β \cdot \frac{2 u sin θ}{g cos β}

u cos θ = \frac{2 u sin θ sin β}{cos β} = 2 u sin θ tan β

cot θ = 2 tan β \Rightarrow θ = cot^{- 1} (2 tan β)

Common Mistakes to Avoid

Watch out

Mixing angles. In projectile motion on level ground, $θ$ is measured from the horizontal. On an incline, $θ$ is measured from the incline's surface, and $α$ (or $β$ ) is the incline's angle from the horizontal. Confusing them wrecks every substitution.
Applying $R = u^{2} sin 2 θ / g$ on an incline. That formula is only for a flat, horizontal ground. On an incline you must redo the analysis in the along/perpendicular frame.
Adding horizontal and vertical velocities as scalars. They are perpendicular components - the speed is $v_{x}^{2} + v_{y}^{2}$ , not $v_{x} + v_{y}$ .
Forgetting horizontal velocity is constant. In vacuum, the horizontal component of a projectile's velocity does not change - the horizontal acceleration is zero.
Using $g$ upward as positive with $u sin θ$ upward as positive. Pick one convention (usually up is +) and stick to it for the whole problem, including at the apex.
Wrong direction of centripetal-like resolution. On an inclined plane, $g cos α$ acts perpendicular to the incline (into it) and $g sin α$ acts along the incline (down the slope). Getting these swapped is a classic error.

Frequently Asked Questions

Q1. Why is the path of a projectile a parabola?

Because horizontal displacement grows linearly with time ( $x = u cos θ \cdot t$ ) while vertical displacement is a quadratic function of time ( $y = u sin θ \cdot t - \frac{1}{2} g t^{2}$ ). Eliminating $t$ gives $y$ as a quadratic function of $x$ - the equation of a parabola.

Q2. Why is $θ = 45°$ the angle for maximum range on level ground?

Because $R = u^{2} sin 2 θ / g$ , and $sin 2 θ$ reaches its maximum value of 1 when $2 θ = 90°$ , i.e., $θ = 45°$ . This maximum range is $R_{ma x} = u^{2} / g$ . On an incline, the optimal angle shifts.

Q3. Do two projectiles fired at complementary angles have equal ranges?

Yes, on a flat horizontal plane. For any angle $θ$ , angles $θ$ and $(90° - θ)$ give the same range because $sin 2 (90° - θ) = sin (180° - 2 θ) = sin 2 θ$ . However, the times of flight and maximum heights are different.

Q4. At the highest point of a projectile's trajectory, what are its velocity and acceleration?

At the apex, the vertical component of velocity is zero, but the horizontal component $u cos θ$ is unchanged. So speed at the apex is $u cos θ$ (not zero). Acceleration is $g$ directed downward - same as everywhere else on the trajectory.

Q5. How does air resistance change projectile motion?

Air resistance reduces both range and maximum height, shortens the time of flight, and makes the trajectory unsymmetrical - the descending portion is steeper than the ascending portion. The exact treatment requires calculus with a velocity-dependent drag term, so board and JEE/NEET problems usually neglect it.

Q6. Why do we set up axes along and perpendicular to the incline, not horizontal and vertical?

Because in the tilted frame, the projectile returns to the surface when perpendicular displacement is zero. That gives a clean equation for time of flight. Also, $g$ is resolved into constant components $g sin α$ and $g cos α$ , which stay constant throughout the motion just like $g$ does in the ordinary frame.

Q7. What is the difference between the position vector and displacement vector?

The position vector $r$ points from the origin to the particle's current location - it depends on the choice of origin. Displacement $Δ r = r_{2} - r_{1}$ is the change in position between two instants - it does not depend on the origin.

Q8. Can projectile motion be treated as two independent 1-D motions?

Yes - and that's the standard trick. Horizontal motion has zero acceleration and constant velocity $u cos θ$ ; vertical motion is uniformly accelerated with acceleration $- g$ and initial vertical velocity $u sin θ$ . Solve each 1-D problem separately, then combine to get position, velocity, and trajectory.

Previous year questions on Motion In Two And Three Dimension

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1. Motion in Two and Three Dimensions

1.1 Vectors and scalars - quick recap

1.2 Scalar and vector products

2. Position and Displacement Vectors

3. Velocity in Two Dimensions

4. Projectile Motion

4.1 Main features of a projectile

4.2 Time of flight, horizontal range, maximum height

4.3 Velocity of the projectile at time t

4.4 Trajectory equation

5. Projectile Motion on an Inclined Plane

5.1 Time of flight (return to inclined surface)

5.2 Maximum height above the inclined surface

5.3 Range along the inclined plane

5.4 Maximum range on the inclined plane

Common Mistakes to Avoid

Frequently Asked Questions

Q1. Why is the path of a projectile a parabola?

Q2. Why is θ=45° the angle for maximum range on level ground?

Q3. Do two projectiles fired at complementary angles have equal ranges?

Q4. At the highest point of a projectile's trajectory, what are its velocity and acceleration?

Q5. How does air resistance change projectile motion?

Q6. Why do we set up axes along and perpendicular to the incline, not horizontal and vertical?

Q7. What is the difference between the position vector and displacement vector?

Q8. Can projectile motion be treated as two independent 1-D motions?

Previous year questions on Motion In Two And Three Dimension

Ready to master Kinematics?

4.3 Velocity of the projectile at time $t$

Q2. Why is $θ = 45°$ the angle for maximum range on level ground?