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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To determine the maximum height of the stone thrown from the bridge, we can use the formula for the vertex of a parabola, which represents the maximum height in this case. For the function \( f(t) = -16t^2 + 35t + 42 \), the time at which it reaches the maximum height can be found using \( t = -\frac{b}{2a} \). Here, \( a = -16 \) and \( b = 35 \), giving us \( t = \frac{35}{32} \). Plugging this back into the height equation provides the maximum height. For the second stone, its initial velocity \( v_0 \) must be such that when it reaches its maximum height, it equals the maximum height of the first stone. By using similar calculations with \( g(t) \) and solving for \( v_0 \) while ensuring the height is the same, we can find the necessary initial velocity that matches the first stone's maximum height. This sets up a fun challenge of balancing two quadratic functions!
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