Solving for 2-Dimensional motion can use the same techniques as 1-Dimension if we divide the problem up into x and y information.
The x information is typically horizontal and the y information is typically vertical. As long as x and y are perpendicular
to each other the events in each direction don't effect each other. They are said to be independent from each other.

It helps to organize our information in an x column and y column with time separate. The x and y columns are completely
independent of each other, but time is shared.

Δt

Δx

Δy

u

u

v

v

a

a

If an object is in free fall we know the gravitational acceleration is purely in the vertical and there is zero horizontal
acceleration.

Play around with this 2-D motion solver to get a feel for what paths an object can take. You can also plug in vlaues to
see if you did a 2-D motion problem correctly.

Example: A ball moving horizontally at 2m/s rolls off a table that is 1.5m high. Find how far the ball travels
horizontally before it hits the ground. Click to rewatch the simulation above.

solution

The information given is organized in the chart below. Vectors directed down or left are negative and vectors directed
up or right are positive.

Δt =

Δx = ?

Δy = -1.5m

u = 2m/s

u = 0

v = 2m/s

v =

a = 0

a = -9.8 m/s²

There isn't enough information to solve the horizontal side. Let start on the vertical.

We know the final velocity is down so after the square root we choose the negative value. Lets plug final velocity back
in and solve for time.

$$v = u + a \Delta t$$ $$-4.98 = 0+(-9.8)\Delta t$$ $$0.51s = \Delta t$$

With this new information we can use time to solve on the horizontal side.

Δt = 0.51s

Δx = ?

Δy = 1.5m

u = 2m/s

u = 0

v = 2m/s

v = -4.98m/s

a = 0

a = 9.8 m/s²

$$\Delta x = u\Delta t + \small\frac{1}{2}a \Delta t^{2}$$ $$\Delta x = 2(0.51) + \small\frac{1}{2}(0)(0.51)^{2}$$ $$\Delta
x = 1.02m$$

Example: A pokéball is thrown at an angle of 60° above the horizon and a speed of 10m/s. If the ball is thrown
from 2.0 meters above the ground how far does the ball go before it misses the charmander and hits the ground. vector reviewsolution

We can throw out the time being zero since that doesn't make sense for the problem.

Δt = 1.93 s

Δx

Δy = 0

u = 5 m/s

u = 8.7 m/s

v

v

a = 0

a = -9.8 m/s²

With the time we can now use horizontal information to solve for the horizontal distance.

$$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$ $$\Delta x = (5)(1.93) + \tfrac{1}{2}(0)(1.93)^{2}$$ $$\Delta x = (5)(1.93)$$
$$\Delta x = 9.65m$$

Δt = 1.93 s

Δx = 9.65 m

Δy = 0

u = 5 m/s

u = 8.7 m/s

v

v

a = 0

a = -9.8 m/s²

Example: A ball is thrown at 35° from the horizon
from ground level
with a speed of 150m/s.
How long will the ball be in the air?
solution

Δt = ?

Δx =

Δy = 0 m

u = 150 cos(35) = 123.0 m/s

u = 150 sin(35) = 86.0 m/s

v =

v =

a = 0

a = -9.8 m/s²

$$\Delta y = u\Delta t + \small\frac{1}{2}a \Delta t^{2}$$ $$0 = (86.0) \Delta t + \small\frac{1}{2}(-9.8)\Delta t^{2}$$

We don't have to use the quadratic equation if we factor out time and solve two separate equations.