Educational

Projectile Motion – Maximum Height

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by

Saim Khalid
in

Introduction

One of the most exciting topics in physics is projectile motion, where an object is launched into the air and follows a curved path under the influence of gravity. From a football kicked across a field to a rocket launched into the sky, projectiles are everywhere in our daily lives.

Projectile motion is a beautiful combination of horizontal and vertical motion happening at the same time. The horizontal component of velocity remains constant, while the vertical component changes due to gravitational acceleration. This blend creates a curved trajectory known as a parabola.

Among the key ideas in projectile motion are the maximum height (how high the projectile goes) and the range (how far it travels horizontally). These concepts are not just classroom physics—they explain how athletes play sports, how engineers design missiles, and how space scientists plan satellite launches.

This article takes a deep dive into projectile motion, especially focusing on maximum height and range. We’ll derive the formulas, explain the physics, give real-life applications, and clear common misconceptions.

Basics of Projectile Motion

Definition

A projectile is any object thrown into space upon which the only force acting is gravity (ignoring air resistance).

Examples:

  • A stone thrown from a cliff
  • A ball kicked at an angle
  • Water sprayed from a hose
  • Bullets fired from a gun

Assumptions

To simplify, we make these assumptions in classical physics:

  1. Air resistance is negligible.
  2. Acceleration due to gravity is constant (g=9.8 m/s2g = 9.8 \, m/s^2g=9.8m/s2).
  3. Motion occurs near the Earth’s surface.

Components of Projectile Motion

When a projectile is launched with velocity uuu at an angle θ\thetaθ:

  • Horizontal component of velocity: ux=ucos⁡θu_x = u \cos \thetaux​=ucosθ
  • Vertical component of velocity: uy=usin⁡θu_y = u \sin \thetauy​=usinθ

These two components act independently:

  • Horizontally: velocity is constant because no horizontal force acts (ignoring air resistance).
  • Vertically: motion is uniformly accelerated due to gravity.

Time of Flight (TTT)

The time of flight is the total time the projectile remains in the air.

Vertical displacement after time TTT = 0 (object returns to same height).

Equation of motion: sy=uyt−12gt2s_y = u_y t – \tfrac{1}{2}gt^2sy​=uy​t−21​gt2

At sy=0s_y = 0sy​=0, t≠0t \neq 0t=0: T=2usin⁡θgT = \frac{2u \sin \theta}{g}T=g2usinθ​

Maximum Height (HHH)

The maximum height is the highest vertical point reached by the projectile.

At maximum height: vertical velocity = 0.

Using first equation of motion: vy=uy−gtv_y = u_y – gtvy​=uy​−gt

At top, vy=0v_y = 0vy​=0: 0=usin⁡θ−gtH0 = u \sin \theta – g t_H0=usinθ−gtH​ tH=usin⁡θgt_H = \frac{u \sin \theta}{g}tH​=gusinθ​

Now, displacement in vertical direction: H=uytH−12gtH2H = u_y t_H – \tfrac{1}{2} g t_H^2H=uy​tH​−21​gtH2​

Substitute: H=(usin⁡θ)(usin⁡θg)−12g(usin⁡θg)2H = (u \sin \theta)\left(\frac{u \sin \theta}{g}\right) – \tfrac{1}{2} g \left(\frac{u \sin \theta}{g}\right)^2H=(usinθ)(gusinθ​)−21​g(gusinθ​)2 H=u2sin⁡2θ2gH = \frac{u^2 \sin^2 \theta}{2g}H=2gu2sin2θ​

So, H=u2sin⁡2θ2g\boxed{H = \frac{u^2 \sin^2 \theta}{2g}}H=2gu2sin2θ​​

Range (RRR)

The range is the horizontal distance traveled by the projectile before it lands.

Range = horizontal velocity × time of flight. R=ux×TR = u_x \times TR=ux​×T R=(ucos⁡θ)(2usin⁡θg)R = (u \cos \theta) \left(\frac{2u \sin \theta}{g}\right)R=(ucosθ)(g2usinθ​) R=u2sin⁡2θgR = \frac{u^2 \sin 2\theta}{g}R=gu2sin2θ​

So, R=u2sin⁡2θg\boxed{R = \frac{u^2 \sin 2\theta}{g}}R=gu2sin2θ​​

Graphical Understanding

  1. Path Shape: A parabola.
  2. Maximum Height: Occurs at the midpoint of time of flight.
  3. Range: The horizontal distance between launch and landing points.
  • At launch: vertical velocity is maximum (usin⁡θu \sin \thetausinθ).
  • At top: vertical velocity = 0, horizontal velocity unchanged.
  • At landing: vertical velocity equals initial downward speed, but opposite in direction.

Conditions for Maximum Range

From formula: R=u2sin⁡2θgR = \frac{u^2 \sin 2\theta}{g}R=gu2sin2θ​

Range depends on sin⁡2θ\sin 2\thetasin2θ, which has a maximum value of 1.

Thus, Rmax=u2gwhen θ=45∘R_{max} = \frac{u^2}{g} \quad \text{when } \theta = 45^\circRmax​=gu2​when θ=45∘

So, a projectile launched at 45° achieves the maximum range.

Relationship Between Height, Range, and Time

  1. Height vs. Time: H=12g(T2)2H = \frac{1}{2} g \left(\frac{T}{2}\right)^2H=21​g(2T​)2
  2. Range vs. Time: R=ucos⁡θ⋅TR = u \cos \theta \cdot TR=ucosθ⋅T
  3. Range and Height: HR=14tan⁡θ\frac{H}{R} = \frac{1}{4} \tan \thetaRH​=41​tanθ

These relationships highlight the deep interconnection between maximum height and horizontal distance.

Real-Life Applications

  1. Sports
    • Basketball shots: Players adjust angle and speed for desired height and distance.
    • Soccer and cricket: Strikers and bowlers rely on projectile motion for perfect throws and kicks.
  2. Engineering
    • Designing cannons, missiles, and rockets involves precise calculation of range.
  3. Military Science
    • Ballistics experts compute projectile trajectories for accuracy.
  4. Water Fountains
    • The height and arc of the water stream follow projectile principles.
  5. Space Exploration
    • Rockets and satellites are launched using extended projectile motion concepts.

Common Misconceptions

  1. “Projectile moves in curved path due to horizontal force.”
    • False. No horizontal force acts (ignoring air resistance). The curve arises from combining constant horizontal velocity and vertical acceleration.
  2. “Heavier projectiles fall faster.”
    • False. Without air resistance, mass does not affect motion—only gravity does.
  3. “Maximum height occurs at maximum range.”
    • Not true. Maximum height occurs at θ=90∘\theta = 90^\circθ=90∘, while maximum range occurs at θ=45∘\theta = 45^\circθ=45∘.

Example Problems

Example 1: Maximum Height

A ball is projected at 20 m/s at 60°. Find maximum height. H=u2sin⁡2θ2gH = \frac{u^2 \sin^2 \theta}{2g}H=2gu2sin2θ​ H=202×sin⁡2602×9.8H = \frac{20^2 \times \sin^2 60}{2 \times 9.8}H=2×9.8202×sin260​ H≈400×0.7519.6≈15.3 mH \approx \frac{400 \times 0.75}{19.6} \approx 15.3 \, mH≈19.6400×0.75​≈15.3m

Example 2: Range

Projectile with velocity 40 m/s at 45°. R=u2sin⁡2θgR = \frac{u^2 \sin 2\theta}{g}R=gu2sin2θ​ R=402sin⁡909.8=16009.8≈163.3 mR = \frac{40^2 \sin 90}{9.8} = \frac{1600}{9.8} \approx 163.3 \, mR=9.8402sin90​=9.81600​≈163.3m

Example 3: Time of Flight

Same projectile as above. T=2usin⁡θgT = \frac{2u \sin \theta}{g}T=g2usinθ​ T=2×40×0.7079.8≈5.77 sT = \frac{2 \times 40 \times 0.707}{9.8} \approx 5.77 \, sT=9.82×40×0.707​≈5.77s

Practice Questions

  1. A projectile is fired at 30 m/s at an angle of 30°. Find its range and maximum height.
  2. At what angle will a projectile travel the same range as another projectile fired at 30° with same speed?
  3. A stone thrown upward at 25 m/s makes an angle of 60° with horizontal. Calculate time of flight.
  4. Prove that for two complementary angles (θ\thetaθ and 90∘−θ90^\circ – \theta90∘−θ), projectiles have equal ranges.

Historical Context

The study of projectile motion began with Galileo Galilei, who proved that projectiles follow parabolic paths. Before Galileo, Aristotle believed projectiles moved in straight lines and then dropped suddenly. Galileo’s experiments with inclined planes and cannonballs laid the foundation of modern kinematics.

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