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A ball is dropped from a bridge at a height of 176.4 over a river. After 2s, a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water simultaneously? The given answer is 24.5 m/s. Help appreciated.
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Ball 1: Initial velocity is zero. Find the time it takes it to fall 176.4 m. Ball 2: Time = time for ball 1 to fall + 2s. Solve for initial velocity of ball 2. For both balls, use:\[\vec{d} = \vec{v}\cdot t + \dfrac{1}{2}\vec{g}t^2\]
Formula didn't show up properly. It should be: \[\textbf{d}= \textbf{v}t + \dfrac{1}{2}\textbf{g}t^2\]
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