https://file.aiquickdraw.com/mathbot/file/1726962671039u7ktmejd.jpg

The problem states that a ball is shot from a cannon with an upward velocity of 40 ft/sec, and the height \( h(t) \) at time \( t \) is modeled by the quadratic equation:

\[
h(t) = -16t^2 + 40t + 1.5
\]

We are tasked with finding the maximum height attained by the ball and graphing the equation.

### 1. Finding the Maximum Height:
The equation is a quadratic function of the form \( h(t) = at^2 + bt + c \), where:
- \( a = -16 \)
- \( b = 40 \)
- \( c = 1.5 \)

Since the coefficient \( a \) is negative, the parabola opens downward, meaning there is a maximum point. The time \( t \) at which the maximum height occurs is given by the formula for the vertex of a parabola:

\[
t_{\text{max}} = \frac{-b}{2a}
\]

Substitute the values for \( a \) and \( b \):

\[
t_{\text{max}} = \frac{-40}{2(-16)} = \frac{40}{32} = 1.25 \text{ seconds}
\]

Now, substitute \( t = 1.25 \) back into the equation for \( h(t) \) to find the maximum height:

\[
h(1.25) = -16(1.25)^2 + 40(1.25) + 1.5
\]
\[
h(1.25) = -16(1.5625) + 50 + 1.5
\]
\[
h(1.25) = -25 + 50 + 1.5 = 26.5 \text{ feet}
\]

### 2. The maximum height attained by the ball is **26.5 feet**.

### 3. Graph of the Equation:
The graph of the function \( h(t) = -16t^2 + 40t + 1.5 \) will be a downward-opening parabola, with the vertex at \( t = 1.25 \) and \( h(1.25) = 26.5 \). The y-intercept occurs at \( t = 0 \), which gives:

\[
h(0) = 1.5 \text{ feet}
\]

Thus, the ball starts at a height of 1.5 feet and reaches its maximum height at 26.5 feet after 1.25 seconds.

Would you like more details, or do you have any questions?

Here are 5 related questions to expand on this problem:
1. How long does it take for the ball to hit the ground after being shot?
2. What is the velocity of the ball when it reaches its maximum height?
3. How would the maximum height change if the initial velocity was 50 ft/sec?
4. What would the trajectory look like if the acceleration due to gravity was different (e.g., on another planet)?
5. How can we use calculus to confirm that the height reaches a maximum at \( t = 1.25 \)?

**Tip:** The time when an object reaches its maximum height in projectile motion can be found by identifying when its velocity becomes zero (first derivative of the height function).

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t) = -16t^2 + 40t + 1.5. Find the maximum height attained by the ball and graph the equation.

Learn how to find the maximum height of a ball shot from a cannon using the quadratic equation h(t) = -16t^2 + 40t + 1.5. This problem illustrates key concepts such as the vertex of a parabola and projectile motion. Suitable for Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation