Motion in a Plane · Lesson 9 of 13

Projectile Height, Flight Time and Range

“Derive the familiar projectile results and establish when maximum range and complementary-angle rules apply.”

Learning Objectives

• Derive rise time, maximum height, total flight time and range. • State the equal-height condition for the standard flight-time and range formulas. • Explain the 45° maximum and complementary launch angles. • Compare high and low trajectories with the same range.

Use the vertical velocity to locate the peak

For launch speed u at angle θ above horizontal, the initial upward velocity is u sinθ. At maximum height the vertical velocity is zero. Set 0 = u sinθ − gt_peak, giving t_peak = u sinθ/g. The maximum height gained above the launch point follows by substituting this time into the vertical position equation, or by using v_y² = u_y² − 2gH with v_y = 0.

Peak time and height gainedLaTeX

t_{peak} = \frac{u sin θ}{g}, H = \frac{u ^{2} sin ^{2} θ}{2 g}

These describe an initially upward component, u sinθ > 0. H is above the release point; maximum height above ground is y₀ + H.

For the time-substitution derivation, u sinθ × (u sinθ/g) gives u²sin²θ/g, and the gravitational term removes half that value. Both methods therefore yield H = u²sin²θ/(2g). At the peak, the remaining horizontal velocity is u cosθ, so the minimum speed during this ideal oblique flight is |u cosθ|.

Return to the launch height

Set y − y₀ = 0 in the vertical equation: t(u sinθ − gt/2) = 0. The first root t = 0 is the launch event. The second root gives the later return to the same height, T = 2u sinθ/g. The flight is symmetric about the peak time: ascent and descent to this same level take equal durations. A different landing height breaks this equal-time result.

Equal-height flight timeLaTeX

T = \frac{2 u sin θ}{g} = 2 t_{peak}

Use this for a later return to the launch level with a positive upward launch component and negligible drag.

Derive the horizontal range

Range R is the horizontal displacement during the chosen complete flight. For equal launch and landing heights, multiply the constant horizontal velocity u cosθ by the flight time. The product is 2u²sinθcosθ/g. Using sin2θ = 2sinθcosθ gives the familiar expression.

Range for equal launch and landing heightsLaTeX

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

This formula assumes a level landing plane through the launch point, constant g and negligible drag. It does not apply unchanged to a cliff launch.

Why 45° gives the maximum in this model

For fixed u and g, only sin2θ can vary in the range formula. Its largest value is one, reached at 2θ = 90° for an upward launch between 0° and 90°. Thus θ = 45° gives R_max = u²/g. A steeper launch increases flight time but reduces horizontal velocity; a shallower launch does the reverse. Their product is optimised at 45° under these conditions.

Maximum level-ground rangeLaTeX

R_{m a x} = \frac{u ^{2}}{g} at θ = 4 5^{\circ}

A height difference, ceiling, drag or another constraint can shift the best launch angle.

Complementary angles give the same range

For angles θ and 90° − θ, the range factors are sin2θ and sin(180° − 2θ), which are equal. Equivalently, angles 45° + α and 45° − α give cos2α in both cases. This proves the complementary-angle result. It requires the same launch speed, the same g and equal release and landing heights.

Two launch angles with equal range— Blue: 30°. Orange: 60°. Both use u = 20 m/s and g = 10 m/s², giving range 34.64 m. The 60° trajectory rises higher and remains airborne longer.

Same range does not mean same flight

Problem
Compare 30° and 60° launches at 20 m/s with g = 10 m/s².

1.For 30°, T = 2 × 20 × 0.5/10 = 2 s and H = 400 × 0.25/20 = 5 m.
2.For 60°, T = 2√3 = 3.464 s and H = 400 × 0.75/20 = 15 m.
3.Both ranges are 400 sin60°/10 = 20√3 = 34.64 m. The lower launch has greater horizontal speed but less flight time.

A complete numerical example

A cricket-ball throw

Problem
A ball is thrown at 28 m/s, 30° above horizontal. Use g = 9.8 m/s². Find peak height, return time and range.

1.The initial vertical component is 28 sin30° = 14 m/s; horizontal component is 14√3 = 24.25 m/s.
2.Peak height is 14²/(2 × 9.8) = 10.0 m. Peak time is 14/9.8 = 1.429 s.
3.Return time is 28/9.8 = 2.857 s, approximately 2.9 s.
4.Range is 28²sin60°/9.8 = 69.28 m, approximately 69 m. Retain extra digits until the final result to avoid unnecessary rounding error.

From maximum range to vertical height

Problem
A person can achieve a maximum level-ground range of 100 m. How high could the same launch speed send a ball vertically upward?

1.Maximum range gives u²/g = 100 m.
2.A vertical throw has H_vertical = u²/(2g) = 50 m.
3.The same launch speed is assumed in both directions. Do not use the height of the 45° trajectory, which would be u²/(4g) = 25 m.

Quick check

For a fixed speed on level ground without drag, which pair gives equal range?

Key Takeaways

• Derive the peak from v_y = 0. • The standard total-flight and range formulas assume a return to launch height. • At fixed speed, 45° maximises ideal level-ground range. • Complementary angles have equal range but different heights and flight times. • Keep height above launch distinct from absolute height above ground.

Previous · Lesson 8

Projectile Motion and the Parabolic Path

Next · Lesson 10

Projectile Problems with Heights and Constraints