Physics 130
West Chester University

Syllabus || Schedule || Problems || Labs

Lesson 5
2D Kinematics

Note: if you see funny symbols, your browser may not be standards compliant, or you may not have the proper fonts. To get these fonts, click here.

Summary of kinematic equations
So you want to solve kinematics problems in two dimensions? If so, you will need the following equations. They are arranged so as to solve for position, but in truth, they can be manipulated algebraically to solve for just about anything in the given equation. For example, you may recall equation 1a) better as v_x = {\Delta x \over \Delta t}.

Position (constant velocity):

1a) \Delta x = v_x \Delta t
1b) \Delta y = v_y \Delta t

Position (using average velocity):
2a) \Delta x = {1 \over 2}(v_{ix} + v_{fx})\Delta t
2b) \Delta y = {1 \over 2}(v_{iy} + v_{fy})\Delta t

Position (constant acceleration, time unknown):
3a) \Delta x = {v_{fx}^2 - v_{ix}^2 \over 2 a_x}
3b) \Delta y = {v_{fy}^2 - v_{iy}^2 \over 2 a_y}

Position (constant acceleration):
4a) \Delta x = v_{ix}\Delta t + {1 \over 2} a_x (\Delta t)^2
4b) \Delta y = v_{iy}\Delta t + {1 \over 2} a_y (\Delta t)^2

Velocity (constant acceleration):
5a) \Delta v_x = a_x \Delta t
5b) \Delta v_y = a_y \Delta t

Our mission today is to start trying to apply these equations to projectile motion. We shall take the approach of using examples to ellucidate these principles.

Example 1: Vertical only

A ball is shot upwards with a velocity of 50 m/s. At what time will it return to Earth?

Solution
We are given one piece of information (the initial velocity vyi and we may infer two other givens (ay = g = -9.8 m/s2) and (vyf = -vyi when it returns to Earth by symmetry. From this point of departure, we must try to determine the appropriate equation to use. Equation 5b) should stand out because it contains all the givens we have and the one unknown. This equation can be rearranged to give:

\Delta t = {\Delta v_y \over a_y}

\Delta t = {v_{yf} - v_{yi} \over a_y}

\Delta t = {-50 m/s - 50 m/s \over -9.8 m/s^2}

\Delta t = 10.2 s

Example 2: Horizontal Launch from a height \Delta y_2 = -20 m
a) Given that a ball has an upwards velocity of 10 m/s and a horizontal velocity of 30 m/s, find the maximum height, \Delta y_1, it reaches.

Solution
First note the implicit givens: a_y = -9.8 m/s^2 and v_{yf} = 0 m/s at the apex. We should focus on equation 3b) as it again gives everything we need.

\Delta y_1 = {(0 m/s)^2 - (10 m/s)^2) \over 2 (-9.8 m/s^2)}

\Delta y_1 = {-100 m^2/s^2 \over -9.8 m/s^2}

\Delta y_1 = 5.10 m

b) Find the time it takes to hit the ground.

Solution
There are many approaches to doing this part of the problem. We shall choose one that is mathematically more challenging but only because the equation we need, 4b), is staring us in the face. We have an initial vertical velocity, we have vertical acceleration, and we know the distance \Delta y_2 = -20 m. To solve this we need to use the quadratic equation. As a review, the quadratic equation is used to solve equations of the form:

Ax^2 + Bx + C = 0

The equation itself is:
x = { - B \pm \sqrt{B^2 - 4AC} \over 2A}

In this case, we must rearrange equation 4b to get:
{1 \over 2}a_y(\Delta t)^2 + v_{yi}\Delta t - \Delta y = 0

We shall set the following variables equal to each other:
A = {1 \over 2} a_y = 0.5(-9.8 m/s^2) = -4.9 m/s^2

B = v_{yi} = 10 m/s

C = -\Delta y = - (-20 m) = 20 m

Thus upon plugging in A, B, and C to the quadratic formula, we get:
\Delta t = {-10 \pm \sqrt{10^2 - 4(-4.9)(20)} \over 2(-4.9)} s

\Delta t = -1.25 s OR 3.28 s

We choose the positive answer of 3.28 seconds from start to finish. While this may look complicated, it saves us from having to combine multiple formula to get an answer.

This portion of the problem can also be done in two steps by using the height obtained in part a) and the known height of the cliff to find the y velocity at the bottom of the cliff. Knowing this final y velocity, we can then use equation 5b) to obtain the flight time.

c) Find the distance the ball travels.

Solution
Having the time of flight from part b), and being given the horizontal velocity: 30 m/s, we can easily find the range using equation 1a).

\Delta x = 30 m/s \times 3.28 s = 98.4 m

Example 3: Given the angle of launch and the net velocity
a) Find the components of velocity given that v = 50.0 m/s and \theta = 30.0^o

Solution
This involves trigonometry. Note that:

v_x = vcos \theta

v_x = 50.0 m/s \times cos(30.0^o) = 43.3 m/s

and
v_y = vsin \theta

v_y = 50.0 m/s \times sin(30.0^o) = 25.0 m/s


b) Find the maximum height
c) Find the time of flight
d) Find the maximum distance

Parts b)-d) can be solved using the methods outlined in Example 2.