You throw a small rock straight up from the edge of a highway bridge that crosses a river. the rock passes you on its way down, 5.00 s after it was thrown. what is the speed of the rock just before it reaches the water 25.0 m below the point where the rock left your hand? ignore air resistance.

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Refer to the diagram shown below. Note that air resistance is ignored, and the acceleration due to gravity is 9.8 m/s². The vertical flight from A to B (the maximum height) and from B to C take equal amounts of time because the flight is parabolic. The time of flight from A to C is given as 5 s, therefore the flight from A to B is 2.5 s. Let u  = vertical launch velocity (m/s). Because the vertical velocity at B is zero, therefore u - (9.8 m/s²)*(2.5 s) = 0 u = 24.5 m/s Let v =  the velocity with which the stone strikes the water. The water surface is at a height of h = -25 m. The equation governing the flight from A to the surface of the water is v² = u² - 2gh v² = (24.5 m/s)² - 2*(9.8 m/s²)*(-25 m)     = 1090.25 v = 33.02 m/s Answer: The velocity with which the stone strikes the water surface is 33.0 m/s (nearest tenth)

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