Projectile motion

Part 1 gave two results for a point that moves under a constant Acceleration $\explain{motion.acceleration}{\mathbf{a}}$ : its Velocity and its Position at any later time. Table 1.2.1 applies both to two worked examples. Gravity is the only cause of a projectile’s acceleration, and near the ground gravity is constant [1]. So a projectile over level ground is such a point, and this part needs no new law of motion. It applies both results of Part 1 to the horizontal and the vertical component of the motion, one component at a time.

This part answers two questions about a projectile launched over level ground. The first chapter derives the range of the projectile: the horizontal distance from the launch point to the landing point. The second chapter finds the launch angle that gives the largest range. This launch angle follows from the range formula alone, so the second chapter needs no new physical assumption.

Part 2 of 22 chapters45–60 min5 numbered equations2 knowledge checks

Projectile Range
Derive and compute the range of a projectile launched over level ground.
20–25 min3 equations1 checkdraft
The Optimal Launch Angle
Show why 45 degrees maximizes a projectile's range for a fixed launch speed, and what that means for other angles.
25–35 min2 equations1 checkdraft

References

Ling et al., University Physics Volume 1 (2016). ch. 4, “Projectile Motion”. https://openstax.org/details/books/university-physics-volume-1 draft ↩

Notation used on this page (5)

\mathbf{a}

Accelerationdraft

The rate at which velocity changes with time. In this course it is held constant over a scenario, so it is the same vector at every instant.

\explain{motion.acceleration}{\mathbf{a}} = \dfrac{d\,\explain{motion.velocity}{\mathbf{v}}}{d\explain{motion.time}{t}}

Units: meters per second squared

v_0

Launch speeddraft

The object's speed at the start of the scenario being analyzed — its initial velocity in a straight-line problem, or its launch speed the instant a projectile leaves the ground.

Units: meters per second

\mathbf{r}

Positiondraft

The object's location at elapsed time t, taken as a vector from a fixed origin so a two-dimensional motion keeps its horizontal and vertical parts separate.

\explain{motion.position}{\mathbf{r}} = (\explain{motion.position.horizontal-coordinate}{x},\ \explain{motion.position.vertical-coordinate}{y})

Symbols: $x$ horizontal coordinate(meters); $y$ vertical coordinate(meters)

Units: meters

t

Timedraft

Elapsed time since the object was launched or released, measured in seconds.

Units: seconds after launch

v(t)

Velocitydraft

The rate at which position changes with time. It is a vector; in a straight-line problem only its size and sign matter, and that is what the scalar v(t) tracks.

\explain{motion.velocity}{\mathbf{v}} = \dfrac{d\,\explain{motion.position}{\mathbf{r}}}{d\explain{motion.time}{t}}\,;\quad \explain{motion.velocity}{v(t)} = \explain{motion.launch-speed}{v_0} + \explain{motion.velocity.constant-acceleration}{a}\,\explain{motion.time}{t}

Symbols: $a$ constant acceleration(meters per second squared)

Units: meters per second