The velocity (in feet/second) of a projectile t seconds after it is launched from a height of 10 feet is given by v(t)= - 15.2t+146. Approximate its height after 3 seconds using 6 rectangles. It is approximatley feet. (Round final answer to nearest tenth. Do NOT round until the final answer.) roj

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Problem Statement:**

The velocity (in feet/second) of a projectile \( t \) seconds after it is launched from a height of 10 feet is given by:
\[ v(t) = -15.2t + 146 \]

Approximate its height after 3 seconds using 6 rectangles.

---

**Solution Steps:**

* The height of the projectile can be found by integrating the velocity function \( v(t) \).

* To approximate this height at \( t = 3 \) seconds, we will use the method of Riemann sums with 6 rectangles. 

* Given the velocity function, \( v(t) = -15.2t + 146 \), over the interval from \( t = 0 \) to \( t = 3 \) seconds, we will divide this interval into 6 subintervals.

* The width of each subinterval (rectangle) is:
  \[ \Delta t = \frac{3 - 0}{6} = 0.5 \text{ seconds} \]

* For each subinterval \( [t_i, t_{i+1}] \), the height of the rectangle can be approximated using the left endpoint \( t_i \):
  \[ \text{Height}_{i} \approx v(t_i) \]

* Summing the areas of these rectangles gives an approximation of the integral of \( v(t) \) over the interval \( [0, 3] \).

---

**Calculation:**

1. Divide the interval \( [0, 3] \) into 6 equal parts:
\[ t = \{0, 0.5, 1, 1.5, 2, 2.5, 3\} \]

2. Calculate \( v(t_i) \) for \( t_i = 0, 0.5, 1, 1.5, 2, 2.5 \):
\[ v(0) = -15.2(0) + 146 = 146 \]
\[ v(0.5) = -15.2(0.5) + 146 = 138.4 \]
\[ v(1) = -15.2(1) + 146 = 130.8 \]
\[ v(1.5) = -15.2(1.5) + 146 = 123.2 \]
Transcribed Image Text:**Problem Statement:** The velocity (in feet/second) of a projectile \( t \) seconds after it is launched from a height of 10 feet is given by: \[ v(t) = -15.2t + 146 \] Approximate its height after 3 seconds using 6 rectangles. --- **Solution Steps:** * The height of the projectile can be found by integrating the velocity function \( v(t) \). * To approximate this height at \( t = 3 \) seconds, we will use the method of Riemann sums with 6 rectangles. * Given the velocity function, \( v(t) = -15.2t + 146 \), over the interval from \( t = 0 \) to \( t = 3 \) seconds, we will divide this interval into 6 subintervals. * The width of each subinterval (rectangle) is: \[ \Delta t = \frac{3 - 0}{6} = 0.5 \text{ seconds} \] * For each subinterval \( [t_i, t_{i+1}] \), the height of the rectangle can be approximated using the left endpoint \( t_i \): \[ \text{Height}_{i} \approx v(t_i) \] * Summing the areas of these rectangles gives an approximation of the integral of \( v(t) \) over the interval \( [0, 3] \). --- **Calculation:** 1. Divide the interval \( [0, 3] \) into 6 equal parts: \[ t = \{0, 0.5, 1, 1.5, 2, 2.5, 3\} \] 2. Calculate \( v(t_i) \) for \( t_i = 0, 0.5, 1, 1.5, 2, 2.5 \): \[ v(0) = -15.2(0) + 146 = 146 \] \[ v(0.5) = -15.2(0.5) + 146 = 138.4 \] \[ v(1) = -15.2(1) + 146 = 130.8 \] \[ v(1.5) = -15.2(1.5) + 146 = 123.2 \]
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