2-D motion
Solving for 2-Dimensional motion uses the same techniques as in 1-D, but with twice the complexity. We will organize our information in an x column and y column with time separate. The vertical and horizontal variables are completely independent of each other, but time is shared. Lets try an example problem.
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.
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 = | |
| x: horizontal (left/right) | y: vertical (up/down) |
|---|---|
| Δx = ? | Δy = -1.5m |
| u = 2m/s | u = 0 |
| v = | v = |
| a = 0 | a = -9.8 m/s² |
There isn't enough information to solve the horizontal side. Let start on the vertical.
$$v^{2} = u^{2}+2a \Delta y$$ $$v^{2} = 0^{2}+2(-9.8)(-1.5)$$ $$v^{2} = 24.9$$ $$v = \sqrt{24.9}$$ $$v = \pm5.42 $$ $$v = -5.42 \small\frac{m}{s}$$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.
$$v = u+a \Delta t$$ $$-5.42 = 0+(-9.8)\Delta t$$ $$0.55s = \Delta t$$
With this new information we can use time to solve on the horizontal side.
| Δt = 0.55s | |
| x: horizontal (left/right) | y: vertical (up/down) |
|---|---|
| Δx = ? | Δy = 1.5m |
| u = 2m/s | u = 0 |
| v = | v = 5.42m/s |
| a = 0 | a = 9.8 m/s² |
Example: A pokeball 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 review
| Δt = ? | |
| x: horizontal | y: vertical |
|---|---|
| Δx = ? | Δy = -2.0m |
| u = 10 cos(60) = 5.0 m/s | u = 10 sin(60) = 8.66 m/s |
| v = ? | v = ? |
| a = 0 | a = -9.8 m/s² |
$$v = u+a \Delta t$$ $$-10.7 = 8.66+(-9.8)\Delta t$$ $$1.97s = \Delta t$$
$$\Delta x = u\Delta t + \small\frac{1}{2}a \Delta t^{2}$$ $$\Delta x = 2(1.97) + \small\frac{1}{2}(0)(1.97)^{2}$$ $$\Delta x = 3.94m$$