91Ó°ÊÓ

When dealing with projectile motion, it's crucial to break down the initial velocity into its horizontal and vertical components. Imagine throwing a coin towards a target; the initial velocity consists of two parts: one that moves the coin horizontally and another that moves it vertically.

The horizontal component, denoted as \(v_{0x}\), is the velocity that facilitates the coin's movement towards the target and is calculated using the formula:

• \(v_{0x} = v_0 \cos \theta\)

where \(v_0\) is the initial speed, and \(\theta\) is the angle of projection. For example, if the initial speed is \(6.4\, \text{m/s}\) and the angle is \(60^\circ\), the horizontal velocity becomes \(3.2\, \text{m/s}\).

Similarly, the vertical component, \(v_{0y}\), represents the velocity propelling the coin upward and is calculated using:

• \(v_{0y} = v_0 \sin \theta\)

For our given values, this works out to \(5.54\, \text{m/s}\). These components are essential for determining other aspects of projectile motion such as time of flight and vertical displacement.

The time of flight is the duration the projectile remains in the air from the moment it is launched until it hits the target or the ground. To calculate this for a projectile like our coin, we focus on the horizontal motion. Since air resistance is ignored, the horizontal velocity remains constant.

We use the equation

• \(x = v_{0x} t\)

to solve for time \(t\). Here, \(x\) is the horizontal distance to the target, which is \(2.1\, \text{m}\) in this problem. Replacing \(v_{0x}\) with our previously calculated \(3.2\, \text{m/s}\), we set up the equation:

• \(2.1 = 3.2t\)
• \(t = \frac{2.1}{3.2} = 0.65625\, \text{s}\)

This time of \(0.65625\, \text{s}\) is critical, as it determines the overall trajectory and helps us calculate the coin's vertical displacement and final vertical velocity.

In projectile motion, vertical displacement tells us how far the projectile travels up or down from its original position. We utilize the time of flight and initial vertical velocity to determine this. The essential formula is:

\[y = v_{0y} t + \frac{1}{2} at^2\]

where:

• \(y\) is the vertical displacement, which represents the shelf height in our example.
• \(v_{0y}\) is the initial vertical velocity we've determined to be \(5.54\, \text{m/s}\).
• \(a\) is the acceleration due to gravity, which is \(-9.8\, \text{m/s}^2\).
• \(t\) is the time of flight we calculated as \(0.65625\, \text{s}\).

Plug the values into our equation:

• \(y = 5.54 \times 0.65625 + \frac{1}{2} (-9.8) \times (0.65625)^2\)
• \(y = 3.63 - 2.11 = 1.52\, \text{m}\)

This result of \(1.52\, \text{m}\) indicates that the shelf is \(1.52\, \text{m}\) above the launch point.

Understanding the vertical velocity of a projectile as it lands is integral to comprehending the projectile's entire motion profile. This measurement indicates how fast the coin is traveling vertically just before it reaches the target.

To calculate the vertical velocity \(v_{y}\) just before landing, you apply the formula:

• \(v_{y} = v_{0y} + at\)

In this situation:

• \(v_{0y}\) again is \(5.54\, \text{m/s}\), the initial vertical velocity.
• \(a\) is \(-9.8\, \text{m/s}^2\), representing the acceleration due to gravity.
• \(t\), our time of flight, is \(0.65625\, \text{s}\).

Putting these into the vertical velocity formula, we get:

• \(v_{y} = 5.54 + (-9.8)(0.65625)\)
• \(v_{y} = 5.54 - 6.43 = -0.89\, \text{m/s}\)

The negative sign in \(-0.89\, \text{m/s}\) shows that, vertically, the coin is moving downward towards the dish at approximately \(0.89\, \text{m/s}\) just before landing.