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To determine the maximum height of the projectile, we need to find the vertex of the quadratic equation modeling its height over time:

\[
h(t) = -16t^2 + 48t + 190
\]

The general form of a quadratic equation is:

\[
h(t) = at^2 + bt + c
\]

For the vertex (which represents the maximum or minimum point of a parabola), the formula for the time \( t \) at which the vertex occurs is:

\[
t = \frac{-b}{2a}
\]

Here, \( a = -16 \) and \( b = 48 \). Plugging these into the formula:

\[
t = \frac{-48}{2(-16)} = \frac{-48}{-32} = 1.5 \, \text{seconds}
\]

Now, we substitute \( t = 1.5 \) into the height equation to find the maximum height:

\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 190
\]
\[
h(1.5) = -16(2.25) + 72 + 190
\]
\[
h(1.5) = -36 + 72 + 190
\]
\[
h(1.5) = 226 \, \text{feet}
\]

Thus, the maximum height of the projectile is **226 feet**.

Let me know if you need further details!

### Related Questions:
1. How would the time to reach the maximum height change if the initial velocity were higher?
2. What would happen to the maximum height if the initial height of the building increased?
3. How do you calculate the total time the projectile remains in the air?
4. How is the acceleration due to gravity represented in the quadratic model?
5. Can you use this method for projectiles launched at angles other than vertical?

### Tip:
The formula \( t = \frac{-b}{2a} \) is useful for finding the vertex of any quadratic function, not just in projectile motion.

A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation h(t) = -16t^2 + 48t + 190. What is the maximum height of the projectile?

Learn how to determine the maximum height of a projectile launched with an initial velocity of 48 feet per second from a 190-foot-tall building using the quadratic equation h(t) = -16t^2 + 48t + 190. This detailed solution includes the application of the vertex formula for quadratic functions. Suitable for students in Grades 9-11.

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