# Chapter 12: Kinematics

## 12.2 2D Kinematics & Projectile motion

A projectile is any object that is thrown or launched. As with 1D kinematics problems, we’re not concerned with what caused the object to be launched, or how it was launched in the first place. All we care about is what happens after it has been launched. At that point, the only force acting on the object is gravity, when air resistance can be neglected.

Now, consider two identical projectiles: one dropped from rest and the other thrown horizontally from the same height, at the same time.

Newton’s first law (section 9.2) tells us that an object’s velocity only changes if there is some force applied to it. For both of the objects, the only force exerted is gravity, which only acts in the vertical direction. Therefore, only the vertical component of a projectile’s velocity changes. The acceleration in the vertical direction is $$g$$. There is zero acceleration in the horizontal direction. (Of course, these statements are only true as long as air resistance can be neglected.) The balls’ trajectories and velocities are shown in the figure below.

For these two objects, we note that the vertical component of each object’s initial velocity is zero, the rate of change of each object’s velocity is the same, and they fall the same vertical distance—in the vertical direction everything is identical!

The vertical and horizontal components of motion are independent from one another. Solving two-dimensional problems is the same as solving one-dimensional problems, except that there is more information to work with. This is where proper organization becomes very important for helping you solve the problem efficiently.

#### Example 12.9

Two kayakers, named Blinky and Clyde, are going down a river when they see a good “jump rocks” spot. This is where you have a tall rock over a deep but calm section of river, which you can climb up and jump off of.

Blinky drops straight down, and Clyde takes a running start.

1. Which kayaker, if either, takes more time to reach the water?
2. Is the splashdown speed (the speed just before hitting the water) of Clyde greater than, equal to, or less than that of Blinky?

Both boaters experience the same acceleration in the vertical direction $$(g)$$, the vertical component of each boater’s initial velocity is the same $$(v_{iy} = 0)$$, and they both fall the same distance—there is no difference at all in their motion in the vertical direction. They take the exact same time to reach the water. Their final velocities also have the same vertical component.

Clyde has some initial velocity in the horizontal direction, so his final velocity will also have some horizontal component. Since his velocity has both $$x$$ and $$y$$ components while Blinky’s velocity only has a $$y$$ component, Clyde’s speed is higher. (You could also answer this part of the question by considering energy conservation.)

#### Example 12.10

A glacier hiker encounters a crevasse. The opposite side of the crevasse is 2.75 m lower than the surface she is currently on, and the crevasse is 4.1 m wide. To cross, she runs straight off the side (i.e., she does not jump up at an angle).

1. What is the minimum speed she needs to cross the crevasse?
2. If her initial speed is 6 m/s, where does she land?
3. If her initial speed is 6 m/s, what is her speed when she lands?

I’ll use a coordinate system where the ledge she is on defines $$x = 0$$, and the height of the lower ledge defines $$y = 0$$. Since she runs off the edge horizontally, her motion is initially only in the $$x$$ direction; there is zero $$y$$ component of the initial velocity. For all three parts of the problem, we know:

\begin{align*} y_i &= 2.75\ \textrm{m} & y_f &= 0 & v_{iy} &= 0 & a_y &= -g \\ x_i &= 0 & & & & & a_x &= 0 \end{align*}

For the first question, we also know $$x_f = 4.1\ \textrm{m}$$, and we are looking for $$v_{ix}$$. Since we have so much more information about the $$y$$ components of motion, we can use them to find the time it takes her to fall from one ledge to the next by using equation (12.1).

\begin{align*} y_f &= y_i + v_{iy}t + \frac{1}{2}a_yt^2 \\ 0 &= y_i + (0)t – \frac{1}{2}gt^2 \\ \hookrightarrow t &= \sqrt{\frac{2y_i}{g}} \end{align*}

With this expression for time, we can determine the speed she needs to have in order to make it to the other side of the crevasse by the time she has fallen to the level of the far side. Use equation (12.1) with horizontal motion components.

\begin{align*} x_f &= x_i + v_{ix}t + \frac{1}{2}a_xt^2 \\ &= 0 + v_{ix}\sqrt{\frac{2y_i}{g}} + \frac{1}{2}(0)t^2 \\ \hookrightarrow v_{ix} &= x_f\sqrt{\frac{g}{2y_i}} \\ &= (4.1\ \textrm{m})\sqrt{\frac{9.81\ \textrm{m/s}^2}{2(2.75\ \textrm{m})}} \\ &= 5.5\ \textrm{m/s} \end{align*}

If she runs any slower than 5.5 m/s, she will have fallen down 2.75 m before traveling the 4.1 m to the side.

For the other two questions, we know $$v_{ix} = 6\ \textrm{m/s}$$, but we do not know $$x_f$$. Since all of the vertical components of motion are the same, we already know the time it takes to reach the other side. Use equation (12.1) to find how far to the side she jumps.

\begin{align*} x_f &= x_i + v_{ix}t + \frac{1}{2}a_xt^2 \\ &= 0 + v_{ix}\sqrt{\frac{2y_i}{g}} + \frac{1}{2}(0)t^2 \\ &= (6\ \textrm{m/s})\sqrt{\frac{2(2.75\ \textrm{m})}{9.81\ \textrm{m/s}^2}} \\ &= 4.5\ \textrm{m} \end{align*}

She lands 40 cm from the edge of the crevasse.

Use equation (12.2) to find the $$x$$ and $$y$$ components of her velocity when she lands.

\begin{align*} v_{fx} &= v_{ix} + a_xt & v_{fy} &= v_{iy} + a_yt \\ &= v_{ix} + (0)t & &= 0 – g\sqrt{\frac{2y_i}{g}} \\ &= 6\ \textrm{m/s} & &= -\sqrt{g^2\left(\frac{2y_i}{g}\right)} \\ & & &= -\sqrt{2gy_i} \\ & & &= -\sqrt{2\left(9.81\ \textrm{m/s}^2\right)(2.75\ \textrm{m})} \\ & & &= -7.3\ \textrm{m/s} \end{align*}

Speed is the magnitude of velocity:

$v_f = \sqrt{v_{fx}^2 + v_{fy}^2} = 9.5\ \textrm{m/s}$

You could also find this by analyzing energy conservation. Initially she has both kinetic energy and gravitational potential energy, and when she lands this has all been converted into kinetic energy. You would find the symbolic solution $$v_f = \sqrt{v_{ix}^2 + 2gy_i}$$, which is exactly the same as what you would find if you substituted the expressions $$v_{fx} = v_{ix}$$ and $$v_{fy} = -\sqrt{2gy_i}$$ from kinematics into $$v_f = \sqrt{v_{fx}^2 + v_{fy}^2}$$.

#### Example 12.11

One of the Highland Games is the hay toss, where competitors use a pitchfork to toss a bale of hay as high as they can. Say one bale leaves the pitchfork 1.8 m above the ground, traveling with a velocity of $$\vec{v} = (1.5\ \textrm{m/s})\hat{x} + (8\ \textrm{m/s})\hat{y}$$. How much time does it take the bale reach its maximum height? What is its maximum height above the ground?

At the maximum height, there is no motion in the $$y$$ direction, so $$v_\textit{fy} = 0$$. I’ll use a $$+y$$ up coordinate system.

We don’t actually need to work with the $$x$$ components of motion at all here! Using the vertical components of motion, we can use equation (12.2) to find the time and equation (12.1) to find the height.

\begin{align*} v_{fy} &= v_{iy} + a_yt & y_f &= y_i + v{iy} t + \frac{1}{2}a_yt^2 \\ 0 &= v_{iy} – gt & &= y_i + v_{iy} t – \frac{1}{2}gt^2 \\ \hookrightarrow t &= \frac{v_{iy}}{g} & &= (1.8\ \textrm{m}) + (8\ \textrm{m/s})(0.82\ \textrm{s}) – \frac{1}{2}\left(9.81\ \textrm{m/s}^2\right)(0.82\ \textrm{s})^2 \\ &= \frac{8\ \textrm{m/s}}{9.81\ \textrm{m/s}^2} & &= 5.1\ \textrm{m} \\ &= 0.82\ \textrm{s} \end{align*}

#### Example 12.12

A long jumper leaves the ground at an angle of 20° to the horizontal, with a speed of 4.5 m/s.

1. How much time does it take her to reach her maximum height?
2. What is her maximum height?
3. What is her total time of flight?
4. How far does she jump?

I’ll split this into two parts: one for the maximum height, and one for the total distance. For both parts I’ll use a $$+y$$ up coordinate system.

##### Part 1: maximum height

At a projectile’s maximum height, the $$y$$ component of the velocity is zero.

For this part of the problem, we only need the vertical components of motion. Given the magnitude and direction of the initial velocity, we can use trig to find the $$y$$ component:

$v_{iy} = v_i\sin\theta$

Use equation (12.2) to find the time and equation (12.1) to find the height.

\begin{align*} v_{fy} &= v_{iy} + a_yt & y_f &= y_i + v_{iy} t + \frac{1}{2}a_yt^2 \\ 0 &= v_i\sin\theta – gt & &= 0 + v_i\sin\theta t – \frac{1}{2}gt^2 \\ \hookrightarrow t &= \frac{v_i\sin\theta}{g} & &= (4.5\ \textrm{m/s})\sin\left(20^\circ\right)(0.16\ \textrm{s}) – \frac{1}{2}\left(9.81\ \textrm{m/s}^2\right)(0.16\ \textrm{s})^2 \\ &= \frac{(4.5\ \textrm{m/s})\sin\left(20^\circ\right)}{9.81\ \textrm{m/s}^2} & &= 0.12\ \textrm{m} \\ &= 0.16\ \textrm{s} \end{align*}

##### Part 2: total distance jumped

We’ll use equation (12.1) to solve both parts of this problem. First we’ll use the $$y$$ components of motion to find the total time of flight, then we’ll use the $$x$$ components of motion to find the distance traveled.

\begin{align*} y_f &= y_i + v_{iy} t + \frac{1}{2}a_yt^2 & x_f &= x_i + v_{ix} t + \frac{1}{2}a_xt^2 \\ 0 &= 0 + v_i\sin\theta t – \frac{1}{2}gt^2 & &= 0 + v_i\cos\theta t + \frac{1}{2}(0)t^2 \\ \hookrightarrow t &= \frac{2v_i\sin\theta}{g} & &= (4.5\ \textrm{m/s})\cos\left(20^\circ\right)(0.31\ \textrm{s}) \\ &= \frac{2(4.5\ \textrm{m/s})\sin\left(20^\circ\right)}{9.81\ \textrm{m/s}^2} & &= 1.3\ \textrm{m} \\ &= 0.31\ \textrm{s} \end{align*}

Look at the symbolic expressions for time and note that the total time of flight is exactly twice the time it takes the person to reach their maximum height. This is the case whenever the initial and final height of a projectile are the same.