Name________________________
Hour _______ Date_____________
Physics
2D Kinematics: Problem Solving- With Angles
Now that you have some background knowledge of 2D kinematics, we will take it one step further and add angles
into the mix.
The first set of equations is 2D kinematics equations you are already familiar with. The second set is the equations
that have angles in them. Notice they look very similar to the ones you already know.
Vertical Motion of a Projectile that falls from rest
Vertical motion of a projectile that has initial velocity
vfy = viy + ayt
vfy2 = viy2 + 2aydy
dy = viyt + ½ ayt2
vy = vi sinϴ
vfy = vi sinϴ + ayt
vfy2 = (visinϴ)2 + 2aydy
dy = (vi sinϴ)t + ½ ayt2
Horizontal motion of a projectile (with angles)
Horizontal Motion of a projectile
vx = vi cosϴ
dx = (vi cosϴ)t
dx = vi2(sin 2ϴ)
ag
vix = vfx = constant
dx = vxt
Time is scalar; meaning it is not direction-dependant.
In this packet, I will show you, step-by-step, how to solve 2D kinematics problems with angles. Then you will
attempt to solve some on your own.
Example 1: A projectile is launched at 25.7 m/s at 23.5o above the ground. Find the y-component of the speed
and the x-component of the speed.
Solution: Think of the situation like this:
y-component
x-component
This is a simple trig problem in the disguise of a 2D kinematics problem.
To solve, use SohCahToa. In 2D problems, cos always goes with the x-component and sin always goes with the ycomponent.
vx = vi cosϴ
vy = vi sinϴ
When you are given a velocity in a 2D problem you are being given the resultant. The equations (given at the
beginning of the packet) break them into the components. That’s why you see the sin and cos in the equations.
Let’s try another one.
Example 2: A golfer hits a golf ball at an angle of 20.0o relative to the ground at a speed of 15.5 m/s. If the golf
ball covers a horizontal distance of 301 m, what is the ball’s maximum height?
Step 1: read the ENTIRE problem.
Step 2: Make a list of your knowns and unknowns. This list should include vi, vf, a, d and t. But it will need to have
all of these quantities listed for both x and y directions.
The ball has initial velocity
so vix = the x component
of the initial velocity.
ax is zero because there is no
force acting in the x direction. No
force = no acceleration.
Given in the problem. dx can
also be called “horizontal
displacement”.
The ball has initial velocity
so viy = the y component
of the initial velocity.
X
vicosϴ
vi
Y
visinϴ
vicosϴ
vf
0
0
a
-9.8 m/s2
301 m
d
?
t
Vfy is 0 because the ball stops
at the top of its path. The top
of its path is the maximum
height.
ay is ALWAYS -9.8 m/s2
because this is the
acceleration due to gravity. It
is always present in the y
direction.
Step 3: Determine what you need to find.
We know that ultimately we need to find dy. It asking this in the question.
Step 4: Choose an equation.
There are 2 equations with dy in them The vest one to use here is vfy2 = (visinϴ)2 + 2aydy.
Step 5: Rearrange the equation.
Step 6: Solve using grids.
In this second example you will fill in the blanks as you go through the steps. The answers are on the very last
page of the packet.
Example 3: Jim throws a ball downward from a rooftop at 7.5 m/s and at angles of 25o from the horizontal.
How fast is the ball moving after 2.0 s?
Step 1: Read the entire problem.
Step 2: Make a list of your knowns and unknowns. This list should include vi, vf, a, d, and t. But it will need to have
all of these quantites listed for both the x and y directions.
X
Y
vi
vf
a
d
t
Step 3: Determine what you need to find.
Ulitimately you need to know ______________________.
Step 4: Choose an equation.
There are 2 equations with _______________ in them.
Which is the better one to use? ________________________
Why? _______________________________________________________.
Step 5: Rearrange the equation:
____________________________
Step 6: Solve using grids.
Now you try one, all on your own. Show your work in the space provided. Answers are on the last page of the
packet.
Example 4: Mary stands at the top of a set of bleachers. She throws a ball at 22.5 m/s and at an angle of 35o
from the horizontal. What is the speed of the ball when it is 15.0 m above the ground? (answer: 21.5 m/s)
Example 5: A ball is launched at an angle of 15o above the ground at a speed of 110 m/s. How long is it in the
air at the highest point in its path? (answer: 2.9 s).
You are ready to start working on 2D Kinematics #3 Worksheet.