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 t seconds after it s thrown is \( f(t)=-16 \mathrm{t}^{2}+35 t+42 \). If a second stone is thrown from the ground, then its height above the ground after \( t \) seconds is given by \( g(t)=-16 \mathrm{t}^{2}+\mathrm{v}_{0} \mathrm{t} \), where \( v_{0} \) 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 velocity will be zero. Therefore, to find the maximum height of the object thrown from the bridge, use the velocity equation, \( v=\frac{\mathrm{ds}}{\mathrm{dt}}=f^{\prime}(t) \). The initial velocity of the second stone would need to be \( \mathrm{v}_{0}=\square \) (Do not round until the final answer. Then round to one decimal place.)

We start by finding the maximum height reached by the stone thrown from the bridge. Its height is given by \[ f(t) = -16t^2 + 35t + 42. \] 1. The velocity is the derivative of \( f(t) \): \[ f'(t) = -32t + 35. \] Setting \( f'(t) = 0 \) gives the time at maximum height: \[ -32t + 35 = 0 \quad \Longrightarrow \quad t = \frac{35}{32}. \] 2. Plug \( t = \frac{35}{32} \) into \( f(t) \) to find 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 \(\left(\frac{35}{32}\right)^2\): \[ \left(\frac{35}{32}\right)^2 = \frac{1225}{1024}. \] Then substitute into the equation: \[ f\left(\frac{35}{32}\right) = -16 \cdot \frac{1225}{1024} + \frac{1225}{32} + 42. \] Simplify the first term: \[ -16 \cdot \frac{1225}{1024} = -\frac{1225}{64}. \] Now express the second term with denominator 64: \[ \frac{1225}{32} = \frac{2450}{64}. \] Thus, we have \[ f\left(\frac{35}{32}\right) = -\frac{1225}{64} + \frac{2450}{64} + 42 = \frac{1225}{64} + 42. \] Express 42 with denominator 64: \[ 42 = \frac{2688}{64}. \] Now, add the fractions: \[ f\left(\frac{35}{32}\right) = \frac{1225 + 2688}{64} = \frac{3913}{64}. \] 3. For the second stone thrown from the ground, its height is given by \[ g(t) = -16t^2 + v_0 t. \] The maximum height for this stone is reached when \( g'(t) = 0 \). Compute the derivative: \[ g'(t) = -32t + v_0. \] Set \( g'(t)=0 \): \[ -32t + v_0 = 0 \quad \Longrightarrow \quad t = \frac{v_0}{32}. \] Then, the maximum height is \[ g\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0}{32}\right)^2 + v_0\left(\frac{v_0}{32}\right). \] Calculate \(\left(\frac{v_0}{32}\right)^2 = \frac{v_0^2}{1024}\). Then, \[ -16\left(\frac{v_0^2}{1024}\right) = -\frac{16v_0^2}{1024} = -\frac{v_0^2}{64}, \] \[ v_0\left(\frac{v_0}{32}\right) = \frac{v_0^2}{32}. \] Thus, \[ g\left(\frac{v_0}{32}\right) = -\frac{v_0^2}{64} + \frac{v_0^2}{32}. \] Write \(\frac{v_0^2}{32}\) as \(\frac{2v_0^2}{64}\): \[ g\left(\frac{v_0}{32}\right) = \frac{-v_0^2 + 2v_0^2}{64} = \frac{v_0^2}{64}. \] 4. To have the same maximum height, set the two heights equal: \[ \frac{v_0^2}{64} = \frac{3913}{64}. \] Multiply both sides by 64: \[ v_0^2 = 3913. \] Taking the positive square root: \[ v_0 = \sqrt{3913}. \] Calculating the numerical value: \[ v_0 \approx 62.6 \, \text{ft/s}. \] Thus, the initial velocity of the second stone must be \( \boxed{62.6} \, \text{ft/s} \).