PHYSICS 1: 2D Kinematics
Kalamazoo Valley Community College
Phy 112
In a 2D Kinematic problem the variables in a
direction must be isolated from the other
direction. Moving Left and Right is not the same
as going Up or Down.
A Vector is any variable which requires a
direction. Displacement, Velocity, and
Acceleration are some common examples.
A Scalar is a variable not associated with a
physical direction. Time, Energy, and Power
Problem Statement
A ping pong ball bounces off a table edge
at 3.30m/s at 55.0°. The table edge is 1.25m
above the ground. Solve for the range of the
ball.
Start by drawing a picture.
If you can’t draw the picture
you will have a very hard time
understanding what comes next.
2
3.30m/s
1
55°
1.25m
Range
3
Build a table
The ball moves in the x and y direction. We will
need to keep of the ball’s x variables separate from
the y variables.
Time is the only value the directions share.
X-direction
Y-direction
Δ𝑥
Δ𝑦
ax
ay
V0x
V0y
Vfx
vfy
time
A ping pong ball bounces off a table edge
at 3.30m/s at 55.0°. The table edge is 1.25m
above the ground. Solve for the range of the
ball.
Remember we need three variables in a direction in order to use
the kinematic equations. The problem statement only states two
numbers.
The additional information needed to solve the problem is part of
knowing the ball is in free-fall. To change velocity requires a push or
a pull in the y-direction this is gravity. In the x-direction what push
or pull will speed up or slow down the ball?
Free-fall: acceleration in the y-direction is 9.80m/s2 pointed down.
Range: total distance travelled by the object in the x-direction.
The ball falls down 1.25m.
We need (-) to say down
X-direction
Y-direction
Δ𝑥
???
Δ𝑦
-1.25m
ax
0
ay
-9.80m/s2
V0x
1.89m/s
V0y
2.70m/s
vfy
???
Vfx
time
???
𝑣𝑜𝑥 = 3.30𝑚
𝑠 ∙ 𝑐𝑜𝑠 55°
Vo=3.30m/s
55°
1
V0x
V0y
𝑣𝑜𝑥 = 1.89𝑚
𝑠
𝑣𝑜𝑦 = 3.30𝑚
𝑠 ∙ 𝑠𝑖𝑛 55°
𝑣𝑜𝑦 = 2.70𝑚
𝑠
The ball is
pulled to the
ground by
gravity.
We need (-)
to say down
There are three variables in the
y-direction and only two in the xdirection. While the problem
statement does not require any
answers for the y-direction, we
need to use the y-direction to
solve for the range.
Δ𝑑 = 𝑣0 ∙ 𝑡 + 12∙a ∙ 𝑡 2
X-direction
Y-direction
Δ𝑥
???
Δ𝑦
-1.25m
ax
0
ay
-9.80m/s2
V0x
1.89m/s
V0y
2.70m/s
vfy
???
Vfx
time
???
𝑣𝑓 = 𝑣0 + a ∙ 𝑡
𝑣𝑓2 = 𝑣02 + 2 ∙ a ∙ Δ𝑑
For the y-direction pick the kinematic equation which allows us to solve for time.
Δ𝑑 = 𝑣0 ∙ 𝑡 + 12∙𝑎 ∙ 𝑡 2
1
𝑚
−1.25𝑚 = 2.70𝑚
∙
𝑡
+
∙ − 9.80 2 ∙ 𝑡 2
𝑠
2
𝑠
Use your calculator’s solve feature or the quadratic equation
to find the two answers.
𝑡 = −0.300𝑠
or
𝑡 = 0.851𝑠
Time is the only variable we can
use across the directions. Why
did we use the positive value?
Now we have three variables in
the x-direction and can solve for
the range.
Δ𝑑 = 𝑣0 ∙ 𝑡 + 12∙a ∙ 𝑡 2
X-direction
Y-direction
Δ𝑥
???
Δ𝑦
-1.25m
ax
0
ay
-9.80m/s2
V0x
1.89m/s
V0y
2.70m/s
vfy
???
Vfx
time
0.851s
𝑣𝑓 = 𝑣0 + a ∙ 𝑡
𝑣𝑓2 = 𝑣02 + 2 ∙ a ∙ Δ𝑑
1
Δ𝑥 = 1.89𝑚
∙
0.851𝑠
+
𝑠
2∙0 ∙ 0.851𝑠
2
Δ𝑥 = 1.61𝑚
𝑅𝑎𝑛𝑔𝑒 = 1.61𝑚
Problem Statement
A basketball player makes a half court shot in
the final seconds of the game. He throws the
ball 1.83m above the floor and tries for the
basket 14.3m away and 3.05m above the
ground. If the ball is released at 7.38m/s @ 50°
does he make the shot?