aspifsGak5u

2021-12-28

If a projectile is fired straight upward from the ground with an initial speed of 128 feet per second, then its height h in feet aftert seconds is given by the function h(t) = -16t + 128t. Find the maximum height of the projectile.

Louis Page

Given height function $h\left(t\right)=-16{t}^{2}+128t$
We have to find the maximum height of the projectile.
$\therefore h\left(t\right)=-16{t}^{2}+128t$
So, ${h}^{\prime }\left(t\right)=-16\left(2t\right)+128$
$⇒{h}^{\prime }\left(t\right)=-32t+128$
To get the critical point, we put ${h}^{\prime }\left(t\right)=0$ and solve for "t".
$\therefore {h}^{\prime }\left(t\right)=0$
$⇒-32t+128=0$
$⇒-32t=-128$
$⇒t=\frac{-128}{-32}$
$⇒t=4$
So, the critical point is $t=4$.
Now, $ht\right)=\frac{d}{dt}\left({h}^{\prime }\left(t\right)\right)=\frac{d}{dt}\left(-32t+128\right)=-32$
So, $ht\right)=-32$
at $t=4,h4\right)=-32<0$
So, by second derivativetest, h(t) has maximum value at $t=4$
$\therefore h\left(4\right)=-16×{4}^{2}+128×4$
$=-256+512$
$=256$
Henca, the maximum height of the projective is 256 feet.

Mary Goodson

$h\left(t\right)=-16{t}^{2}+64t$
complete the square
$h\left(t\right)=-16\left({t}^{2}-4t+4\right)+64$
$h\left(t\right)=-16{\left(t-2\right)}^{2}+64$
This is an equation of a parabola that opens upward with vertex at (2,64) maximum height of the projectile $=64$ ft (2 sec after it is fired)

user_27qwe

For this case we have the following function:
Deriving the function we have:
h'(t)=-32t + 128
Equaling zero we have:
-32t + 128 = 0
We clear the time:
$t=\frac{128}{32}$
t = 4
Substituting in the height equation:
$h\left(4\right)=-16\ast \left(4{\right)}^{2}+128\ast \left(4\right)$
h(4) = 256 feet
The maximum height is:
h (4) = 256 feet

Mr Solver

The given function is:
$h\left(t\right)=-16t+128t$
To find the vertex, we first need to convert the function to vertex form. The vertex form of a quadratic function is given by:
$f\left(t\right)=a\left(t-h{\right)}^{2}+k$
where $\left(h,k\right)$ represents the coordinates of the vertex.
Let's begin by factoring out the common factor of $t$ from the equation:
$h\left(t\right)=t\left(-16+128\right)$
Simplifying further:
$h\left(t\right)=112t$
Now, let's rewrite the equation by completing the square. We can do this by adding and subtracting the square of half the coefficient of $t$:
$h\left(t\right)=112\left(t-0{\right)}^{2}$
Comparing this with the vertex form equation, we can see that $h=0$ and $k=0$. Therefore, the vertex of the parabola is $\left(0,0\right)$.
Since the projectile is fired straight upward, the maximum height is attained at the vertex of the parabolic function. In this case, the maximum height is 0 feet.
Hence, the maximum height of the projectile is 0 feet.

To find the vertex, we can use the formula $h=-\frac{b}{2a}$, where $a$ is the coefficient of the squared term and $b$ is the coefficient of the linear term.
In this case, $a=-16$ and $b=128$. Substituting these values into the formula, we get:
$h=-\frac{128}{2\left(-16\right)}=-\frac{128}{-32}=4$
Therefore, the $t$-coordinate of the vertex is $t=4$. To find the corresponding height, we substitute this value back into the equation:
$h\left(4\right)=-16\left(4{\right)}^{2}+128\left(4\right)$
Simplifying this expression gives us:
$h\left(4\right)=-16\left(16\right)+128\left(4\right)=-256+512=256$
The maximum height of the projectile is 256 feet.

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