Answer :
Sure, let me walk you through how to understand the height [tex]\( h \)[/tex] of the ball as a function of time [tex]\( t \)[/tex]. We start with the equation:
[tex]\[ h = 3 \sin \left(\frac{\pi}{2}(t + 2)\right) + 5 \][/tex]
This equation represents a sinusoidal function, which describes oscillatory motion. Let's dissect this step-by-step:
### 1. Understand the Components of the Equation:
- Amplitude: The amplitude of the sine function is [tex]\( 3 \)[/tex]. This determines how far the maximum and minimum values of the height, [tex]\( h \)[/tex], deviate from the middle value.
- Frequency: The sine function has a coefficient [tex]\(\frac{\pi}{2}\)[/tex] inside the argument, which affects the period of the oscillation. Recall that the period [tex]\( T \)[/tex] of the sine function [tex]\( \sin(k t) \)[/tex] is calculated as [tex]\( T = \frac{2\pi}{k} \)[/tex]. Here, [tex]\( k = \frac{\pi}{2} \)[/tex], so the period [tex]\( T \)[/tex] is:
[tex]\[ T = \frac{2\pi}{\frac{\pi}{2}} = 4 \][/tex]
- Horizontal Shift: The term [tex]\( t + 2 \)[/tex] indicates a horizontal shift to the left by 2 units.
- Vertical Shift: The +5 indicates that the entire sine wave is shifted upwards by 5 units.
### 2. Graph Characteristics:
The general shape of the graph will be a sine wave oscillating between a minimum height of [tex]\( 2 \)[/tex] feet and a maximum height of [tex]\( 8 \)[/tex] feet (since [tex]\( 5 - 3 = 2 \)[/tex] and [tex]\( 5 + 3 = 8 \)[/tex]) with these points repeating every 4 seconds due to the period. The sine wave hits the midline at [tex]\( 5 \)[/tex] feet.
### 3. Starting and Key Points:
Given the horizontal shift, the starting point of the [tex]\( t \)[/tex]-values should be at [tex]\( t = -2 \)[/tex] when [tex]\( h \)[/tex] is at the midline value of [tex]\( 5 \)[/tex] feet. Key points occur where the sine function reaches [tex]\( \frac{\pi}{2} \)[/tex], [tex]\( \pi \)[/tex], [tex]\( \frac{3\pi}{2} \)[/tex], etc. This equates to the sine wave starting at [tex]\( t = -2 \)[/tex]*and achieving a maximum value at [tex]\( t = 0 \)[/tex].
### 4. Detailed Values:
To achieve a detailed understanding, let's look at some specific values:
- At [tex]\( t = -2 \)[/tex]:
[tex]\[ h = 3 \sin(\frac{\pi}{2} * 0) + 5 = 3 \sin(0) + 5 = 5 \][/tex]
- At [tex]\( t = -1 \)[/tex]:
[tex]\[ h = 3 \sin(\frac{\pi}{2} 1) + 5 = 3 \sin(\frac{\pi}{2}) + 5 = 3 1 + 5 = 8 \][/tex]
- At [tex]\( t = 0 \)[/tex]:
[tex]\[ h = 3 \sin(\frac{\pi}{2} 2) + 5 = 3 \sin(\pi) + 5 = 3 0 + 5 = 5 \][/tex]
- At [tex]\( t = 1 \)[/tex]:
[tex]\[ h = 3 \sin(\frac{\pi}{2} 3) + 5 = 3 \sin(\frac{3\pi}{2}) + 5 = 3 (-1) + 5 = 2 \][/tex]
- At [tex]\( t = 2 \)[/tex]:
[tex]\[ h = 3 \sin(\frac{\pi}{2} 4) + 5 = 3 \sin(2\pi) + 5 = 3 0 + 5 = 5 \][/tex]
### 5. Graph Representation:
With these values and additional ones following the same sinusoidal nature, you can sketch the graph. Here’s what it should look like:
- The curve starts at [tex]\( t = -2 \)[/tex] at [tex]\( h = 5 \)[/tex].
- Rises to [tex]\( h = 8 \)[/tex] at [tex]\( t = -1 \)[/tex].
- Returns to [tex]\( h = 5 \)[/tex] at [tex]\( t = 0 \)[/tex].
- Drops to [tex]\( h = 2 \)[/tex] at [tex]\( t = 1 \)[/tex].
- Rises back to [tex]\( h = 5 \)[/tex] at [tex]\( t = 2 \)[/tex].
This pattern repeats every 4 seconds, creating a sinusoidal wave oscillating between 2 and 8 feet.
### Conclusion:
The graph of the function [tex]\( h = 3 \sin \left(\frac{\pi}{2}(t+2)\right)+5 \)[/tex] would showcase a periodic behavior with the described properties. Each cycle lasting 4 seconds, shifted left by 2 units, and vertically shifted up by 5 units.
[tex]\[ h = 3 \sin \left(\frac{\pi}{2}(t + 2)\right) + 5 \][/tex]
This equation represents a sinusoidal function, which describes oscillatory motion. Let's dissect this step-by-step:
### 1. Understand the Components of the Equation:
- Amplitude: The amplitude of the sine function is [tex]\( 3 \)[/tex]. This determines how far the maximum and minimum values of the height, [tex]\( h \)[/tex], deviate from the middle value.
- Frequency: The sine function has a coefficient [tex]\(\frac{\pi}{2}\)[/tex] inside the argument, which affects the period of the oscillation. Recall that the period [tex]\( T \)[/tex] of the sine function [tex]\( \sin(k t) \)[/tex] is calculated as [tex]\( T = \frac{2\pi}{k} \)[/tex]. Here, [tex]\( k = \frac{\pi}{2} \)[/tex], so the period [tex]\( T \)[/tex] is:
[tex]\[ T = \frac{2\pi}{\frac{\pi}{2}} = 4 \][/tex]
- Horizontal Shift: The term [tex]\( t + 2 \)[/tex] indicates a horizontal shift to the left by 2 units.
- Vertical Shift: The +5 indicates that the entire sine wave is shifted upwards by 5 units.
### 2. Graph Characteristics:
The general shape of the graph will be a sine wave oscillating between a minimum height of [tex]\( 2 \)[/tex] feet and a maximum height of [tex]\( 8 \)[/tex] feet (since [tex]\( 5 - 3 = 2 \)[/tex] and [tex]\( 5 + 3 = 8 \)[/tex]) with these points repeating every 4 seconds due to the period. The sine wave hits the midline at [tex]\( 5 \)[/tex] feet.
### 3. Starting and Key Points:
Given the horizontal shift, the starting point of the [tex]\( t \)[/tex]-values should be at [tex]\( t = -2 \)[/tex] when [tex]\( h \)[/tex] is at the midline value of [tex]\( 5 \)[/tex] feet. Key points occur where the sine function reaches [tex]\( \frac{\pi}{2} \)[/tex], [tex]\( \pi \)[/tex], [tex]\( \frac{3\pi}{2} \)[/tex], etc. This equates to the sine wave starting at [tex]\( t = -2 \)[/tex]*and achieving a maximum value at [tex]\( t = 0 \)[/tex].
### 4. Detailed Values:
To achieve a detailed understanding, let's look at some specific values:
- At [tex]\( t = -2 \)[/tex]:
[tex]\[ h = 3 \sin(\frac{\pi}{2} * 0) + 5 = 3 \sin(0) + 5 = 5 \][/tex]
- At [tex]\( t = -1 \)[/tex]:
[tex]\[ h = 3 \sin(\frac{\pi}{2} 1) + 5 = 3 \sin(\frac{\pi}{2}) + 5 = 3 1 + 5 = 8 \][/tex]
- At [tex]\( t = 0 \)[/tex]:
[tex]\[ h = 3 \sin(\frac{\pi}{2} 2) + 5 = 3 \sin(\pi) + 5 = 3 0 + 5 = 5 \][/tex]
- At [tex]\( t = 1 \)[/tex]:
[tex]\[ h = 3 \sin(\frac{\pi}{2} 3) + 5 = 3 \sin(\frac{3\pi}{2}) + 5 = 3 (-1) + 5 = 2 \][/tex]
- At [tex]\( t = 2 \)[/tex]:
[tex]\[ h = 3 \sin(\frac{\pi}{2} 4) + 5 = 3 \sin(2\pi) + 5 = 3 0 + 5 = 5 \][/tex]
### 5. Graph Representation:
With these values and additional ones following the same sinusoidal nature, you can sketch the graph. Here’s what it should look like:
- The curve starts at [tex]\( t = -2 \)[/tex] at [tex]\( h = 5 \)[/tex].
- Rises to [tex]\( h = 8 \)[/tex] at [tex]\( t = -1 \)[/tex].
- Returns to [tex]\( h = 5 \)[/tex] at [tex]\( t = 0 \)[/tex].
- Drops to [tex]\( h = 2 \)[/tex] at [tex]\( t = 1 \)[/tex].
- Rises back to [tex]\( h = 5 \)[/tex] at [tex]\( t = 2 \)[/tex].
This pattern repeats every 4 seconds, creating a sinusoidal wave oscillating between 2 and 8 feet.
### Conclusion:
The graph of the function [tex]\( h = 3 \sin \left(\frac{\pi}{2}(t+2)\right)+5 \)[/tex] would showcase a periodic behavior with the described properties. Each cycle lasting 4 seconds, shifted left by 2 units, and vertically shifted up by 5 units.
