Projectile Motion: Range, Height and the 45° Rule

Why a thrown object follows a curved path

A ball launched upward and forward continues to move horizontally while gravity changes its vertical velocity. These two motions happen simultaneously. Treating them separately turns an apparently complicated curve into two simple equations. Projectile motion is a model, not a claim that real objects encounter no air.

What you will learn

Resolve launch velocity into horizontal and vertical components.
Predict position, time of flight, maximum height and horizontal range.
Explain why 45° gives maximum range only for equal launch and landing heights without air resistance.
Test predictions against the free Projectile Motion Lab.

The model and its assumptions

Choose a fixed launch speed v₀, launch angle θ above horizontal, and gravitational acceleration g. Use horizontal x and upward y as the coordinate directions. The ideal model assumes constant downward gravitational acceleration, no aerodynamic drag and a level landing surface at the same elevation as launch. The projectile is treated as a point mass. For numerical illustrations here, g = 9.81 m/s².

Resolve the initial velocity

v₀ₓ = v₀ cos θ ; v₀ᵧ = v₀ sin θ

Gravity has no horizontal component in this model. Therefore horizontal velocity remains v₀ cos θ. Vertically, gravity continually reduces upward velocity, brings it momentarily to zero at the top, and then increases its downward magnitude. The horizontal and vertical velocities combine to give the changing direction of motion.

Position at any instant

x(t) = (v₀ cos θ)t

y(t) = (v₀ sin θ)t − ½gt²

vᵧ(t) = v₀ sin θ − gt

Eliminate time from the first two equations to obtain y = x tan θ − gx²/(2v₀² cos² θ), a parabola when the assumptions are satisfied. Notice that constant horizontal speed does not mean constant overall speed.

Deriving flight time, peak height and range

At the highest point, vertical velocity is zero. Hence the time to reach the top is tₚ = v₀ sin θ/g. The trajectory is symmetric about the peak when it starts and lands at equal heights, so total flight time is twice tₚ. Substituting the peak time into y(t) gives the maximum height. Multiplying constant horizontal velocity by flight time gives the range.

T = 2v₀ sin θ/g

H = v₀² sin² θ/(2g)

R = v₀² sin(2θ)/g

These formulas are for equal launch and landing elevations only. They must not be used unchanged for a projectile launched from a roof or a cliff.

Worked numerical example: 24 m/s at 30°

Launch a projectile at 24 m/s at 30°, and use g = 9.81 m/s². Horizontal speed is 24 cos 30° = 20.78 m/s; initial vertical speed is 24 sin 30° = 12.0 m/s. Flight time T = 24/9.81 = 2.45 s. Maximum height H = 144/(19.62) = 7.34 m. Horizontal range R = 576 sin 60°/9.81 = 50.85 m. Units are essential: range and height are metres, time is seconds.

Why 30° and 60° give the same range

The range equation contains sin(2θ). For θ = 30°, sin 60° = 0.866; for θ = 60°, sin 120° is also 0.866. Both produce the same 50.85 m range at the same speed. However, the 60° shot rises higher and remains airborne longer. At θ = 45°, sin 90° = 1, so the ideal maximum range at 24 m/s is approximately 58.72 m. Complementary angles produce equal range, not identical trajectories.

Guided investigation: predict, change, explain

Open the Projectile Motion Lab and use the launch-speed and angle controls.
Set 24 m/s and 30°; record predicted T, H and R before starting.
Try 60° at the same speed. Record which readouts match and which change.
Set 45°. Compare the range to the other two trials.
Keep the angle fixed and double launch speed where the controls permit. Predict the multiplier for range and height.
Change only one variable per trial and record why the model and your estimate might differ by rounding.

Misconceptions to avoid

“Gravity starts acting only after the projectile reaches the top.” Gravity acts throughout the flight.
“Horizontal acceleration becomes negative near landing.” In a no-drag model, horizontal acceleration stays zero.
“The maximum-range angle is always 45°.” That is conditional on equal launch and landing heights and no air resistance.
“The path is realistic for a feather or parachute.” Air drag strongly alters these trajectories.

Practice and solutions

A 20 m/s projectile is launched at 45° on level ground with g = 10 m/s². Find R. Answer: R = 20²/10 = 40 m.
At fixed speed, which has longer flight time: 30° or 60°? Answer: 60°, because T is proportional to sin θ.
If launch speed becomes three times as large while angle and g remain fixed, what happens to R and H? Answer: both become nine times as large in this ideal model.
At the highest point, is the projectile at rest? Answer: normally no; vertical velocity is zero but horizontal velocity remains nonzero.

Classroom and exam connection

For NCERT Class 11, the essential skill is separating perpendicular components of motion. For JEE questions, identify the launch and landing heights before applying a memorised range formula. State assumptions, draw axes, and use consistent SI units. The same reasoning supports sports trajectories and the initial motion of a water jet, but real examples may require a drag model.

Continue your investigation

Open the free simulation linked below, compare calculated and displayed quantities, and use its printable worksheet. A teacher can ask students to submit a prediction table before revealing readouts. This approach values explanation over button-clicking.

Frequently asked questions

Is 45° always the best launch angle?

Only for ideal no-drag motion with equal launch and landing heights at a fixed speed.

What happens to horizontal velocity?

It remains constant in the ideal model; gravity accelerates vertically.

Why do complementary angles have equal range?

Their doubled angles have equal sine values in R = v₀² sin(2θ)/g.

Does doubling launch speed double the range?

No. At fixed angle, the ideal level-ground range becomes four times as large.