At what launch angle is the horizontal range of a projectile on level ground maximized?

45°. Range depends on (2 ), which reaches its maximum value of 1 when 2 = 90^ , hence = 45^ .

Which launch angle will give the exact same horizontal range as an angle of 25°?

65°. Complementary angles have identical ranges: 90^ - 25^ = 65^ .

If air resistance is neglected, what happens to the horizontal component of velocity during flight?

It remains strictly constant.. Because there is no horizontal force acting on the projectile, a_x = 0 and v_x remains constant throughout.

If the initial velocity of a projectile is doubled, what happens to its maximum height H?

It quadruples (4x). Maximum height is proportional to v_0^2. Doubling velocity increases height by (2)^2 = 4 times.

A projectile is launched from ground level at an angle . At the peak of its trajectory, what are its velocity and acceleration vectors?

Velocity is horizontal (v_0 ); acceleration is g downward. 1. At the peak, vertical velocity v_y = 0, but the horizontal component v_x = v_0 remains unchanged. 2. Gravity acts downward continuously, so acceleration is constant at g downwards at all points during flight.

Physics · Class 11-12 · AP Physics 1

Projectile Motion and Kinematics

Projectile motionis the motion of an object projected into the air, moving solely under the influence of gravity along a curved parabolic path, demonstrating independence of horizontal and vertical motions.

1Independence of Motions & Trajectory Diagram

Galileo first demonstrated that projectile motion can be split into twomutually independent motions: uniform motion at constant velocity horizontally, and uniformly accelerated motion vertically due to gravity. Below is the complete trajectory with velocity vectors at every stage of flight.

Projectile Trajectory and Velocity Vector Components: Launch Point (0,0), Initial Components, Ascending Flight, Apex / Maximum Height H, Descending Flight, Total Range R — Projectile Trajectory and Velocity Vector Components
A classical parabolic trajectory launched at angle theta. Notice that the horizontal velocity vector vx remains constant in magnitude throughout flight, while vertical velocity vy decreases to 0 at the peak and reverses downward.
1Launch Point (0,0) — Initial speed v0 directed at launch angle theta above horizontal
2Initial Components — v0x = v0 cos(theta) and v0y = v0 sin(theta)
3Ascending Flight — Vertical velocity vy reduces due to downward gravitational acceleration g
4Apex / Maximum Height H — At the top, vy = 0 instantly; speed is strictly equal to vx
5Descending Flight — vy points downward and increases in magnitude until impact
6Total Range R — The horizontal distance travelled when the projectile returns to launch height

Horizontal vs. Vertical Kinematic Components

Dimension	Acceleration	Velocity Behavior	Governing Equation
Horizontal (x)	$a_x = 0$	Constant: $v_x = v_0 \cos\theta$	$x = (v_0 \cos\theta) \cdot t$
Vertical (y)	$a_y = -g = -9.8\,\text{m/s}^2$	Changes linearly: $v_y = v_0 \sin\theta - gt$	$y = (v_0 \sin\theta)t - \frac{1}{2}gt^2$

Horizontal (x)
Acceleration
$a_x = 0$
Velocity Behavior
Constant: $v_x = v_0 \cos\theta$
Governing Equation
$x = (v_0 \cos\theta) \cdot t$
Vertical (y)
Acceleration
$a_y = -g = -9.8\,\text{m/s}^2$
Velocity Behavior
Changes linearly: $v_y = v_0 \sin\theta - gt$
Governing Equation
$y = (v_0 \sin\theta)t - \frac{1}{2}gt^2$

2Derivations for Time of Flight, Height, and Range

Derivation — Maximum Height H and Time of Flight T

1
$v_y = v_{0y} - gt = v_0\sin\theta - gt$
Vertical velocity equation
2
$0 = v_0\sin\theta - gt_{up} \implies t_{up} = \frac{v_0\sin\theta}{g}$
$At \text{highest point}\ , vy = 0$
3
$T = 2 t_{up} = \frac{2v_0\sin\theta}{g}$
Total flight time for level ground
4
$H = \frac{(v_0\sin\theta)^2}{2g} = \frac{v_0^2\sin^2\theta}{2g}$
$\text{Max height using}\ v^2 - u^2 = 2as$
5
$R = v_{0x} \cdot T = (v_0\cos\theta)\left(\frac{2v_0\sin\theta}{g}\right) = \frac{v_0^2\sin(2\theta)}{g}$
Horizontal range using 2sin(theta)cos(theta) = sin(2theta)

3Key Formulas & Quick Reference Table

Standard Projectile Formulas (Level Ground)

Quantity	Symbol	Formula	Condition for Maximum
Time of Flight	$T$	$T = \frac{2v_0 \sin\theta}{g}$	Maximized at $\theta = 90^\circ$ (vertical throw)
Maximum Height	$H$	$H = \frac{v_0^2 \sin^2\theta}{2g}$	Maximized at $\theta = 90^\circ$
Horizontal Range	$R$	$R = \frac{v_0^2 \sin 2\theta}{g}$	Maximized at $\theta = 45^\circ$ (where $\sin 90^\circ = 1$ )

Time of Flight
Symbol
$T$
Formula
$T = \frac{2v_0 \sin\theta}{g}$
Condition for Maximum
Maximized at $\theta = 90^\circ$ (vertical throw)
Maximum Height
Symbol
$H$
Formula
$H = \frac{v_0^2 \sin^2\theta}{2g}$
Condition for Maximum
Maximized at $\theta = 90^\circ$
Horizontal Range
Symbol
$R$
Formula
$R = \frac{v_0^2 \sin 2\theta}{g}$
Condition for Maximum
Maximized at $\theta = 45^\circ$ (where $\sin 90^\circ = 1$ )

Exam tip: Complementary Angles Rule: Two launch angles that sum to

90^\circ

(like

30^\circ

and

60^\circ

, or

15^\circ

and

75^\circ

) achieve the EXACT SAME horizontal range

R

, because

\sin(2(90^\circ - \theta)) = \sin(180^\circ - 2\theta) = \sin(2\theta)

Common mistake: Common Trap: Saying "the acceleration at the highest point is zero". This is FALSE! At the peak, velocity is horizontal (

v_y = 0

, but

v_x \neq 0

), and acceleration remains constant at

g = 9.8\,\text{m/s}^2

downwards!

4Worked Problem Step by Step

Real-life example · Worked Numerical Problem

A football is kicked with an initial speed of $v_0 = 20\,\text{m/s}$ at an angle of $\theta = 30^\circ$ to horizontal (take $g = 10\,\text{m/s}^2$ ).

1. Initial velocities:
$v_{0x} = 20 \cos 30^\circ = 20 \times \frac{\sqrt{3}}{2} = 17.32\,\text{m/s}$
$v_{0y} = 20 \sin 30^\circ = 20 \times 0.5 = 10\,\text{m/s}$

2. Time of flight:
$T = \frac{2(10)}{10} = 2\,\text{seconds}$

3. Maximum height:
$H = \frac{10^2}{2(10)} = \frac{100}{20} = 5\,\text{meters}$

4. Range:
$R = v_{0x} \cdot T = 17.32 \times 2 = 34.64\,\text{meters}$ .

Exam question · AP Physics 1 / JEE Main

A projectile is launched from ground level at an angle

\theta

. At the peak of its trajectory, what are its velocity and acceleration vectors?

A.Velocity is zero; acceleration is zero.
B.Velocity is horizontal ( $v_0 \cos\theta$ ); acceleration is $g$ downward.
C.Velocity is vertical; acceleration is horizontal.
D.Velocity is $v_0$ ; acceleration is zero.

Remember

Projectile motion is a superposition of constant velocity horizontal motion and constant acceleration vertical motion.
Horizontal acceleration is zero ( $a_x = 0$ ); vertical acceleration is constant downwards ( $a_y = -g$ ).
At the highest point, vertical velocity is zero ( $v_y = 0$ ), but horizontal velocity is non-zero ( $v_x = v_0 \cos\theta$ ).
The range is identical for complementary angles $\theta$ and $90^\circ - \theta$ .
Maximum horizontal range occurs at launch angle $\theta = 45^\circ$ .

Check yourself

Question 1
At what launch angle is the horizontal range of a projectile on level ground maximized?
Question 2
Which launch angle will give the exact same horizontal range as an angle of 25°?
Question 3
If air resistance is neglected, what happens to the horizontal component of velocity during flight?
Question 4
If the initial velocity of a projectile is doubled, what happens to its maximum height $H$ ?