Set up the equation 15 = 20 − 5 t 2 to find the time when the height is 15 m, which gives t = 1 second.
Set up the equation 10 = 20 − 5 t 2 to find the time when the height is 10 m, which gives t = 2 seconds.
The time interval during which the camera should film the person is between 1 second and 2 seconds.
The interval is [ 1 , 2 ] .
Explanation
Understanding the Problem We are given the height of a stuntperson as a function of time: h = 20 − 5 t 2 . We need to find the time interval when the stuntperson is between 15 m and 10 m above the ground. This means we need to solve the inequality 10 ≤ h ≤ 15 .
Finding the Time at 15 m First, let's find the time when the stuntperson is at 15 m. We set h = 15 and solve for t :
15 = 20 − 5 t 2 5 t 2 = 20 − 15 5 t 2 = 5 t 2 = 1 t = ± 1 Since time cannot be negative, we have t = 1 second.
Finding the Time at 10 m Next, let's find the time when the stuntperson is at 10 m. We set h = 10 and solve for t :
10 = 20 − 5 t 2 5 t 2 = 20 − 10 5 t 2 = 10 t 2 = 2 t = ± 2 Since time cannot be negative, we have t = 2 ≈ 1.414 seconds.
Determining the Time Interval Therefore, the camera should film the person between 1 second and 2 seconds. In interval notation, this is [ 1 , 2 ] .
Examples
Understanding the time interval during which an object falls within a certain height range is crucial in various real-world applications. For instance, in film production, knowing the precise timing for capturing a stunt is essential for safety and visual impact. Similarly, in engineering, calculating the time an object spends within a specific altitude can help in designing safety measures or optimizing performance. This problem demonstrates how mathematical models can provide valuable insights into dynamic processes.
The camera should record the stuntperson between 1 second and 2 seconds, which is approximately 1.414 seconds. This is the time interval when the height is between 15 m and 10 m above the ground, consistent with the given height equation. Thus, the filming interval is [ 1 , 2 ] .
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