3. Projectile Motion

Overview

Two dimensional projectile motion can be broken down into horizontal and vertical components. The horizontal component experiences zero acceleration. The vertical component experiences acceleration due to gravity.

Slides

3.1 Horizontal Projectile Motion

Skills and Understanding

Apply kinematic equations to solve projectile motion problems in 2D.
Use trigonometry to find lengths of triangles corresponding to magnitudes of vectors.
Understand that projectile motion can be decomposed into horizontal and vertical motion, and these can be analyzed separately.
Understand that vertical acceleration due to gravity is independent of horizontal motion.

Equations

\[ \begin{array}{cc} v = \frac{\Delta x}{\Delta t} \quad \quad \quad \quad & v_{f} = v_{i} + a \cdot \Delta t \\ a = \frac{\Delta v}{\Delta t} \quad \quad \quad \quad & \Delta x = \frac{1}{2} ( v_{i} + v_{f} ) \Delta t \\ v_{avg} = \frac{v_{i}+v_{f}}{2} \quad \quad \quad \quad & \Delta x = v_{i} \Delta t + \frac{1}{2} a \Delta t ^{2} \\ \quad \quad \quad \quad & v_{f}^{2} = v_{i}^{2} + 2 a \Delta x \end{array} \]

Vocabulary

Acceleration \(a\) is the rate at which velocity changes. It is calculated as \(a = \frac{\Delta v}{\Delta t}\)
Displacement is the change in position from the start to the end of a path. Given an initial position \(x_{i}\) and final position \(x_{f}\), the displacement \(\Delta x\) is calculated as \(\Delta x = x_{f} - x_{i}\)
Freefall describes when an object is falling due to gravity alone.
A projectile is an object with some initial velocity that is accelerating due to gravity.
Velocity is the rate at which position is changing. It is calculated as \(v = \frac{\Delta x}{\Delta t}\)

3.1 Horizontal Projectile

In projectile motion, we have a two dimensional problem which we can break into horizontal and vertical components. Importantly, there is no acceleration in the horizontal direction while the acceleration in the vertical direction is due to gravity.

A horizontal projectile is a special case because the initial vertical velocity is zero, or \[ v_{yi} = 0 \text{ m/s}\] This is useful because it can make our algebra much easier! Note that we are now using subscripts to distinguish between horizontal \(x\) motion and vertical \(y\) motion.

A ball is rolling at 3.2 m/s when it rolls off the edge of a 1.2 m tall table. How far from the table does the ball land?

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We break this problem into two pieces: vertical, and horizontal.

Vertical

Considering only the vertical motion of the ball as it falls, \[ v_{yi} = 0 \text{ m/s} \] \[ \Delta y = -1.2 \text{ m} \] \[ a_{y} = -10 \text {m/s}^{2} \] With this information we can find the amount of time \(\Delta t\) that the ball will be falling before it hits the ground. Using \[ \Delta y = v_{yi} \Delta t + \frac{1}{2} a_{y} \Delta t ^{2} \] We plug in and solve to get \[ \Delta t = \sqrt{ \frac{2 \Delta y}{a_{y}} } = \sqrt{\frac{2 (-1.2)}{(-10)}} = 0.49 \text{ s}\]

Horizontal

Now that we know \(\Delta t\), we can consider the horizontal motion. Using \[ \Delta x = v_{x} \Delta t \] we plug in to get \[ \Delta x = (3.2)(0.49) = 1.57 \text{ m} \]

3.2 Vector Components

Given a vector \(v\) with magnitude \(|v|\) and direction \(\theta\) measured from the positive \(x\)-axis, the honrizontal and vertical components can be calculated with \[v_{x} = v \cdot \cos(\theta)\] \[v_{y} = v \cdot \sin(\theta)\] Consider the vector in the graph below. Try dragging the tip of the vector to change its magnitude (length) and direction. Explore how changing the vector changes its horizontal component (red) and vertical component (blue).

3.3 Projectile at an Angle

Play with the PhET projetile simulator. What do you notice? What do you wonder?

A marble is launched from ground height with an initial velocity \(v_{i}=5.7\) m/s at an angle of \(72^{\circ}\). How far horizontally does the marble travel before coming back to the ground?

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First we find the horizontal and vertical components of the velocity. \[v_{x} = 5.7 \cdot \cos(72^{\circ}) \approx 1.761 \] \[v_{yi} = 5.7 \cdot \sin(72^{\circ}) \approx 5.421 \] We then break the problem into two: horizontal and vertical.

Vertical

We can use \(v_{yi}\) to find how much time the marble spends in the air. One way to do this is with \[ a_{y} = \frac{v_{yf}-v_{yi}}{\Delta t} \] Given that \(a_{y}=-10\), \(v_{yi} = 5.421\), and \(v_{yf}=-5.421\) because of symmetry, we can solve for \(\Delta t\) to get \[ \Delta t = \frac{-5.421 - 5.421}{-10} = 1.084 \text{ s} \]

Horizontal

Using the value of \(\Delta t\) from the vertical portion, we can find the horizontal displacement: \[ \Delta x = v_{x} \Delta t = (1.761)(1.084) = 1.9 \text{ m} \]

3.4 Projectile at a Height

Previously, we analyzed a projectile that landed at the same height from which it was launched. Let's consider the case where a projectile lands at a different height.

Our useful strategy is to break problems into smaller problems.

A projectile is launched from a height of 3.4 m with an initial velocity \(v_{i}=12.5\) m/s at an angle of \(51^{\circ}\). How far horizontally does the marble travel before landing on the ground?

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First we find the horizontal and vertical components of the velocity. \[v_{x} = 12.5 \cdot \cos(51^{\circ}) \approx 7.867 \] \[v_{yi} = 12.5 \cdot \sin(51^{\circ}) \approx 9.714 \] We will find the time \(\Delta t_{1}\) from start to peak, and time \(\Delta t_{2}\) from peak to ground. Then with the total time we can find the horizontal displacement.

From start to peak

Using \(a=\frac{v_{f}-v_{i}}{\Delta t}\) we can find the time to peak with \[\Delta t_{1} = \frac{0 - 9.714}{-10} = 0.9714\] For that amount of time, our projectile traveled upwards a distance of \[ \Delta y = v_{yi} \Delta t_{1} + \frac{1}{2} a \Delta t_{1}^{2} = 4.718 \text{ m} \] which we add to 3.4 to get a peak height of \(8.118\) m.

From peak to ground

Using the peak height, we can find the time to the ground, \(\Delta t_{2}\), \[ \Delta t_{2} = \sqrt{\frac{2\cdot(-8.118)}{-10}} = 1.274 \] and the horizontal displacement from start to end is \[ \Delta x = v_{xi} \cdot (\Delta t_{1} + \Delta t_{2}) = 17.7 \text{ m} \]