Worked Examples

Example 1: Soccer Ball Kicked at an Angle

A soccer player kicks a ball with an initial velocity of 20 m/s at an angle of 30° above the horizontal. Find the range, maximum height, and time of flight. (Use g = 10 m/s²)

Given: v₀ = 20 m/s, θ = 30°, g = 10 m/s²

sin 30° = 0.5, cos 30° ≈ 0.866, sin 60° ≈ 0.866

Time of Flight: T = (2 × v₀ × sin θ) / g = (2 × 20 × 0.5) / 10 = 20 / 10 = 2.0 s

Maximum Height: H = (v₀² × sin²θ) / (2g) = (400 × 0.25) / 20 = 100 / 20 = 5.0 m

Range: R = (v₀² × sin 2θ) / g = (400 × sin 60°) / 10 = (400 × 0.866) / 10 = 346.4 / 10 = 34.6 m

Example 2: Ball Thrown Horizontally from a Cliff

A ball is thrown horizontally at 15 m/s from the edge of a cliff 45 m high. How far from the base of the cliff does it land? (Use g = 10 m/s²)

Given: v₀ = 15 m/s, θ = 0°, H = 45 m, g = 10 m/s²

Time of Flight (free fall from height): H = ½ × g × T² → 45 = ½ × 10 × T² → T² = 9 → T = 3.0 s

Horizontal Range: R = v₀ × T = 15 × 3.0 = 45.0 m

The ball lands 45 m from the base of the cliff.

Example 3: Projectile Launched at 45° for Maximum Range

A cannonball is fired at 40 m/s at an angle of 45°. Find the maximum range and the maximum height reached. (Use g = 10 m/s²)

Given: v₀ = 40 m/s, θ = 45°, g = 10 m/s²

sin 45° ≈ 0.707, sin 90° = 1

Range: R = (v₀² × sin 2θ) / g = (1600 × 1) / 10 = 160 m

Time of Flight: T = (2 × v₀ × sin θ) / g = (2 × 40 × 0.707) / 10 = 56.56 / 10 = 5.66 s

Maximum Height: H = (v₀² × sin²θ) / (2g) = (1600 × 0.5) / 20 = 800 / 20 = 40 m

Common Mistakes to Avoid

• Using the full initial velocity instead of its vertical component for height calculations. The maximum height depends only on the vertical component (v₀ sin θ), not the full speed. Always resolve the velocity into horizontal and vertical components before substituting into any formula.
• Forgetting that horizontal velocity is constant throughout the flight. Many students apply deceleration to the horizontal direction. Gravity acts only vertically; horizontal velocity does not change (ignoring air resistance).
• Assuming maximum range always occurs at 45° regardless of launch height. The 45° rule applies only when the projectile lands at the same height from which it was launched. For launches from a cliff or elevated surface, the optimal angle shifts below 45°.
• Confusing time to reach maximum height with total time of flight. The time to reach peak height is T/2, not T. Total time of flight is twice the time to reach maximum height only when launch and landing heights are equal.
• Using degrees in a calculator set to radians (or vice versa). This produces wildly incorrect values for sin and cos. Always verify your calculator mode before computing trigonometric functions in physics problems.

Real-World Applications

• Sports biomechanics — basketball free throw: Coaches and sports scientists use projectile motion equations to determine the optimal release angle (typically around 45–52°) and speed for a free throw, maximizing the chance of the ball passing cleanly through the hoop.
• Military and forensic ballistics: Artillery crews calculate the launch angle and muzzle velocity needed to hit a target at a known horizontal distance. Forensic investigators use the same principles to reconstruct the trajectory of a projectile from a crime scene.
• Aerospace — rocket and missile trajectory planning: During the unpowered phase of flight, rockets and re-entry vehicles follow projectile paths. Engineers compute range and time of flight to plan landing zones and recovery operations.
• Irrigation and firefighting — water jet reach: Civil engineers and firefighters calculate the range of a water jet by treating it as a projectile. Adjusting the nozzle angle to near 45° maximizes horizontal reach, which is critical when fighting fires from a distance or designing sprinkler systems.

Frequently Asked Questions

Why is the range maximum at a launch angle of 45°?

The range formula is R = (v₀² × sin 2θ) / g. The term sin 2θ reaches its maximum value of 1 when 2θ = 90°, which means θ = 45°. At any other angle, sin 2θ is less than 1, so the range is shorter. This assumes launch and landing are at the same height.

What happens to the horizontal velocity during projectile motion?

Horizontal velocity remains constant throughout the entire flight because no horizontal force acts on the projectile (gravity is purely vertical). This means the horizontal component of velocity at launch equals the horizontal component at landing.

At what point during flight is the speed of a projectile the lowest?

Speed is lowest at the maximum height. At that instant, the vertical component of velocity is zero, so only the horizontal component (v₀ cos θ) remains. Since horizontal velocity never changes, this is the minimum total speed during the flight.

Does a heavier object travel farther than a lighter one when launched at the same angle and speed?

No. In the absence of air resistance, mass does not appear in any of the projectile motion equations. Both objects experience the same gravitational acceleration (g ≈ 9.8 m/s²) regardless of mass, so they follow identical trajectories.

How does air resistance affect projectile motion?

Air resistance reduces both range and maximum height by opposing the motion in the direction of travel. It also means the optimal launch angle for maximum range drops below 45°. The standard projectile equations assume no air resistance, which is a valid approximation for dense, slow-moving objects over short distances.

Can two different launch angles give the same range?

Yes. Complementary angles — pairs that add up to 90° — produce the same range when launched from the same height with the same speed. For example, launches at 30° and 60° yield identical ranges because sin(2 × 30°) = sin 60° = sin(2 × 60°) = sin 120°, and sin 60° = sin 120°.