We appreciate your visit to A projectile is launched from a building 190 feet tall with an initial velocity of 48 feet per second The path of the projectile is. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
We are given the height equation for the projectile:
[tex]$$
h(t) = -16t^2 + 48t + 190,
$$[/tex]
where [tex]$t$[/tex] is the time in seconds and [tex]$h(t)$[/tex] is the height in feet.
Step 1. Find the Time at Maximum Height
For a quadratic function of the form
[tex]$$
h(t) = at^2 + bt + c,
$$[/tex]
the time at which the maximum (or minimum) height occurs is found at the vertex of the parabola. This occurs at
[tex]$$
t_{\text{max}} = -\frac{b}{2a}.
$$[/tex]
Plugging in the values [tex]$a = -16$[/tex] and [tex]$b = 48$[/tex], we have:
[tex]$$
t_{\text{max}} = -\frac{48}{2(-16)} = -\frac{48}{-32} = 1.5 \text{ seconds}.
$$[/tex]
Step 2. Calculate the Maximum Height
Now substitute [tex]$t = 1.5$[/tex] into the height equation:
[tex]$$
\begin{aligned}
h(1.5) &= -16(1.5)^2 + 48(1.5) + 190 \\
&= -16(2.25) + 72 + 190 \\
&= -36 + 72 + 190 \\
&= 226 \text{ feet}.
\end{aligned}
$$[/tex]
Conclusion
The maximum height reached by the projectile is
[tex]$$
\boxed{226 \text{ feet}}.
$$[/tex]
[tex]$$
h(t) = -16t^2 + 48t + 190,
$$[/tex]
where [tex]$t$[/tex] is the time in seconds and [tex]$h(t)$[/tex] is the height in feet.
Step 1. Find the Time at Maximum Height
For a quadratic function of the form
[tex]$$
h(t) = at^2 + bt + c,
$$[/tex]
the time at which the maximum (or minimum) height occurs is found at the vertex of the parabola. This occurs at
[tex]$$
t_{\text{max}} = -\frac{b}{2a}.
$$[/tex]
Plugging in the values [tex]$a = -16$[/tex] and [tex]$b = 48$[/tex], we have:
[tex]$$
t_{\text{max}} = -\frac{48}{2(-16)} = -\frac{48}{-32} = 1.5 \text{ seconds}.
$$[/tex]
Step 2. Calculate the Maximum Height
Now substitute [tex]$t = 1.5$[/tex] into the height equation:
[tex]$$
\begin{aligned}
h(1.5) &= -16(1.5)^2 + 48(1.5) + 190 \\
&= -16(2.25) + 72 + 190 \\
&= -36 + 72 + 190 \\
&= 226 \text{ feet}.
\end{aligned}
$$[/tex]
Conclusion
The maximum height reached by the projectile is
[tex]$$
\boxed{226 \text{ feet}}.
$$[/tex]
Thanks for taking the time to read A projectile is launched from a building 190 feet tall with an initial velocity of 48 feet per second The path of the projectile is. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
