A stone is thrown with an initial velocity of $$ 35 \mathrm{ft} / \mathrm{s} $$ from the edge of a bridge that is 42 ft above the ground. The height of this stone above the ground f seconds after it is thrown is $$ f(t)=-16 \mathrm{t}^{2}+35 \mathrm{t}+42 $$. If a second stone is thrown from the ground, then its height above the ground after t seconds is given by $$ g(\mathrm{t})=-16 \mathrm{t}^{2}+\mathrm{v}_{0} \mathrm{t} $$, where $$ \mathrm{v}_{0} $$ is the initial velocity of the second stone. Determine the value of $$ \mathrm{v}_{0} $$ such that the two stones reach the same maximum height. When an object that is thrown upwards reaches its highest point (just before it starts to fall back to the ground), its maximum height of the object thrown from the bridge, use the

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We are given the height functions

$$
f(t) = -16t^2 + 35t + 42 \quad \text{and} \quad g(t) = -16t^2 + v_0 t.
$$

We want the maximum heights of the two stones to be the same.

**Step 1. Find the maximum height of the stone thrown from the bridge**

The function $$ f(t) = -16t^2 + 35t + 42 $$ is a downward-opening parabola. Its maximum occurs at the vertex. The time at the vertex is found by

$$
t = -\frac{b}{2a},
$$

where $$ a = -16 $$ and $$ b = 35 $$. Thus,

$$
t = -\frac{35}{2(-16)} = \frac{35}{32}.
$$

Now substitute $$ t = \frac{35}{32} $$ into $$ f(t) $$ to get the maximum height:

$$
f\left(\frac{35}{32}\right) = -16\left(\frac{35}{32}\right)^2 + 35\left(\frac{35}{32}\right) + 42.
$$

First, compute

$$
\left(\frac{35}{32}\right)^2 = \frac{1225}{1024}.
$$

Then,

$$
-16 \left(\frac{1225}{1024}\right) = \frac{-19600}{1024}.
$$

Next, compute

$$
35\left(\frac{35}{32}\right) = \frac{1225}{32} = \frac{1225 \times 32}{1024} = \frac{39200}{1024}.
$$

Finally, convert $$ 42 $$ to a fraction with denominator 1024:

$$
42 = \frac{42 \times 1024}{1024} = \frac{43008}{1024}.
$$

Now, add the three parts:

$$
f\left(\frac{35}{32}\right) = \frac{-19600 + 39200 + 43008}{1024} = \frac{62608}{1024}.
$$

Simplify the fraction by dividing numerator and denominator by 16:

$$
\frac{62608 \div 16}{1024 \div 16} = \frac{3913}{64}.
$$

So the maximum height of the first stone is

$$
H = \frac{3913}{64} \quad (\text{ft}).
$$

**Step 2. Find the maximum height of the stone thrown from the ground**

For the second stone, the height function is

$$
g(t) = -16t^2 + v_0 t.
$$

This parabola has its vertex at

$$
t = -\frac{b}{2a} = -\frac{v_0}{2(-16)} = \frac{v_0}{32}.
$$

Substitute $$ t = \frac{v_0}{32} $$ into $$ g(t) $$:

$$
g\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0}{32}\right)^2 + v_0\left(\frac{v_0}{32}\right).
$$

Compute the first term:

$$
\left(\frac{v_0}{32}\right)^2 = \frac{v_0^2}{1024} \quad \Rightarrow \quad -16\left(\frac{v_0^2}{1024}\right) = -\frac{16v_0^2}{1024} = -\frac{v_0^2}{64}.
$$

And the second term is

$$
v_0\left(\frac{v_0}{32}\right) = \frac{v_0^2}{32}.
$$

Thus,

$$
g\left(\frac{v_0}{32}\right) = -\frac{v_0^2}{64} + \frac{v_0^2}{32}.
$$

To combine the fractions, write $$ \frac{v_0^2}{32} $$ as $$ \frac{2v_0^2}{64} $$:

$$
g\left(\frac{v_0}{32}\right) = -\frac{v_0^2}{64} + \frac{2v_0^2}{64} = \frac{v_0^2}{64}.
$$

Thus, the maximum height of the second stone is

$$
\frac{v_0^2}{64}.
$$

**Step 3. Equate the maximum heights**

Set the maximum heights equal:

$$
\frac{v_0^2}{64} = \frac{3913}{64}.
$$

Multiply both sides by 64:

$$
v_0^2 = 3913.
$$

Taking the positive square root (since $$ v_0 $$ is a speed):

$$
v_0 = \sqrt{3913}.
$$

Thus, the value of $$ v_0 $$ such that the two stones reach the same maximum height is

$$
v_0 = \sqrt{3913}.
$$

This is approximately

$$
v_0 \approx 62.57 \, \text{ft/s}.
$$
The value of $$ v_0 $$ is $$ \sqrt{3913} $$ ft/s, which is approximately 62.57 ft/s.

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