Maximum Height, Time of Flight, and Range

Expert reviewed • 22 November 2024 • 7 minute read

What is the Maximum Height?

The maximum height of a projectile is the highest point it reaches during its flight. At this point, the vertical component of the projectile’s velocity is momentarily zero.

What are the key Factors Affecting Maximum Height?

Initial launch height
Initial vertical velocity

Notably, the initial horizontal velocity does not affect the maximum height.

How do you Calculate the Maximum Height of a Projectile?

To calculate the maximum height, we use the equation:

v_y^2 = u_y^2 + 2a_yΔy

Where:

$v_y$ is the final vertical velocity (0 at maximum height)
$u_y$ is the initial vertical velocity
$a_y$ is the vertical acceleration (usually -9.8 m/s² due to gravity)
$\Delta y$ is the change in height

At maximum height, $v_y=0$ , so we can rearrange the equation to solve for $\Delta y$ :

\Delta y = -\frac{u_y^2}{2a_y}

What is Time of Flight?

The time of flight is the total duration of the projectile’s motion, from launch to landing. It is primarily affected by the initial vertical velocity and is independent of the horizontal velocity.

Calculating Time of Flight

For projectiles launched from and landing at the same height, we can use the equation:

y = u_y t + \frac{1}{2}a_y t^2

Where:

$y$ is the vertical displacement (0 for same start and end heights)
$u_y$ is the initial vertical velocity
$t$ is the time of flight
$a_y$ is the vertical acceleration (-9.8 m/s²)

Solving for t gives us two solutions: $t=0$ (the start of motion) and $t = \frac{2u_y}{g}$ (the time of flight).

What is Range?

The range of a projectile is the horizontal distance it travels from launch to landing. It depends on both the initial horizontal velocity and the time of flight.

Calculating Range

The range can be calculated using the equation:

R = u_xt

Where:

$R$ is the range
$u_x$ is the initial horizontal velocity
$t$ is the time of flight

Practice Question 1

A ball is launched with an initial velocity of 50 m/s at an angle of 30° above the horizontal. Calculate:

The maximum height reached
The time of flight
The range of the projectile

</examplequestion

Summary Points

Maximum height depends on initial vertical velocity and is reached when vertical velocity becomes zero.
Time of flight is determined by initial vertical velocity and is independent of horizontal velocity.
Range depends on both initial horizontal velocity and time of flight.
Projectile motion can be analysed by separating it into horizontal and vertical components.
Horizontal velocity remains constant throughout the motion (ignoring air resistance).
Vertical motion is subject to constant acceleration due to gravity.

Understanding these principles allows for accurate prediction and analysis of projectile motion in various scenarios, from sports to ballistics and beyond.

First, let’s break down the initial velocity into its components:

u_x = 50cos (30°) = 43.3 m/s \\u_y = 50sin (30°) = 25 m/s

Maximum height:

\Delta y = -\frac{u_y^2}{2a_y} \\= -\frac{25^2}{2(-9.8 )} \\= 31.9m

Time of flight:

t = \frac{2u_y}{g} \\= \frac{2(25 )}{9.8 } \\= 5.1s

Range:

R = u_xt \\= (43.3)(5.1 ) \\ = 220.8 m

Initial Velocity and Launch Angle

Up Next

Optimal Angle for Maximum Projectile Range

Return to Module 5: Advanced Mechanics