**Example**

While biking, you keep your velocity at a constant 12 m/s magnitude directed east with respect to the earth. The wind starts blowing toward the west with a speed of 5 m/s, with respect to the earth. What is your resultant speed with respect to the earth at the time the wind starts blowing.

**Solution**

We will draw a diagram to show the two velocities. The resultant velocity is the vector sum of the two velocities considered in the problem. We will consider east to be the positive *x*-axis direction (see Figure 4.8).

Hence, the resultant speed is:

**v**_{r} = **v**_{1} – **v**_{2}

**v**_{r} = **v**_{1} – **v**_{2}

**v**_{r} = 12 – 5 = 7 m/s

**v**_{r} is 7 m/s in the east direction.

**Projectile Motion**

A classic case of two-dimensional motion is *projectile motion*. Imagine you throw an object horizontally, as in skipping a rock. In a gravity-free and friction-free world, the rock will move horizontally until it interacts with an obstacle as shown in Figure 4.9. The diagram shows the constant velocity (magnitude and direction).

If we consider the friction with the air, the rock will use energy through conversion to heat during its path and it will slow down. Figure 4.10 shows the velocity and the acceleration (constant acceleration).

If we consider no friction but an acceleration downward, as in the case of the acceleration due to gravity, *g* = 9.8 m/s^{2}, what happens to the velocity? You might surmise the answer from everyday happenings, but this is a good place to use what we learned and to learn something new. The graph in Figure 4.11 shows the object moving down, and then acceleration due to gravity is also down. As we have seen before, the acceleration is in the same direction as the change in velocity, hence the change in velocity, **Δv**, has to be also downward. But **Δv** = **v**_{2} – **v**_{1} and that means according to the diagram that **v**_{2} has to be the same direction as **v**_{1} and larger than **v**_{1}.

Hence, we will complete the diagram in Figure 4.11 with the vector velocity at other positions (see Figure 4.12).

What will happen if the object moves up? Then the acceleration is opposite to the initial velocity, and we have a case similar to the previous one with friction. The object will slow down.

In the previous discussion, we separated the horizontal and vertical motion, and we examined different motion. What happens if we consider a more complex motion on both *x* and *y* direction? We will be able then to add the *x* and *y* components of displacement, velocity, and acceleration and then fully characterize the two-dimensional motion: We obtain *projectile motion*.

**Example**

A rock is thrown with a speed of 10 m/s at an angle of 30 degrees from the horizontal. Draw a diagram of motion and show in a few places, both when the rock is ascending and descending, the velocity, and the acceleration.

**Solution**

*v*_{0} = 10 m/s

α = 30°

We will consider a two-dimensional frame and a friction-free motion.

On the horizontal *x*-axis, the object moves with constant velocity (*v*_{x} = constant and *v*_{x} = 0 m/s^{2}) because there is no friction and no other interaction. Also, if the object does not start from the origin of the reference frame, an offset has to be considered, *x*_{0}:

*x* = *x*_{0} + *v _{x}* ·

**Δ**

*t*

and *v _{x}* is the component of the speed on the

*x*-axis (see Figure 4.13).

In the vertical direction, the object moves up decelerating (slowing down), stops at the top of the trajectory, and then returns down accelerating (see Figure 4.14). All this motion is determined by the constant acceleration due to gravity.

The initial components of the velocity can be found from the data in Figure 4.15.

*v*_{0x} = *v*_{0} · cos α

*v*_{0y} = *v*_{0} · sin α

With all this information, the *x*-*y* graphical description of the previous motion is shown in Figure 4.16.

The maximum height (vertical) and distance (horizontal) the object reaches is based on the initial speed and the angle of throw. The horizontal distance for a symmetric projectile motion as in this previous example (the projectile starts and finishes at the same *y* position) is called *range*. To summarize the equation of motion for the *x*- and *y*-axis are:

### x-axis

*v*_{0x} = *v*_{0} · cos α = constant

*x* = *x*_{0} + *v*_{0} · cos α Δ*t*

*a*_{x} = 0 m/s^{2}

### y-axis

*v*_{0y} = *v*_{0} · sin α

*y* = *y*_{0} + *v*_{0y} · Δ*t* + *a _{y}* · Δ

*t*

^{2}/2

*a*_{y} = constant

If we eliminate the time dependence from both *x*(*t*) and *y*(*t*), then we get the *y*(*x*) dependence:

And we can determine the *range* and the *maximum height* to be:

In these equations, the initial position is considered to be the origin:

*x*_{0} = 0

*y*_{0} = 0

The time it takes the object to describe this trajectory is called *time of flight* and can be calculated to be:

Figure 4.17 shows the different paths an object takes for a fixed speed with different throwing angles. The longest path (maximum range at a given speed) is for the 45°.

Practice problems of this concept can be found at: Motion in Two Dimensions Practice Questions

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