The Bridge and the Rock Using energy considerations and assuming negligible air resistance, find the speed of a rock just before it hits the water if it is thrown from a bridge 16.4 m above water with an initial speed of 16.4 m/s? (Verify that the speed is independent of the direction the rock is thrown.)

24.30 m/s

Explanation:

If the stone is thrown horizontally, our horizontal velocity u₁ = 16.4 m/s and its initial vertical velocity, U = 0 and its final vertical velocity, V is thus

V² = u² + 2gs where s = height of bridge above water = 16.4 m, g = 9.8 m/s²

v² = 0 + 2 * 9.8 * 16.4 = 321.44

v = √321.44 = 17.93 m/s

Since the horizontal velocity is constant, our resultant velocity just before it hits the water is v₁ = √ (u₁² + v²) = √ (16.4² + 17.93²) = √ (268.96 + 321.44) = √590.4 = 24.30 m/s

If the stone is thrown vertically, its initial vertical velocity, u = 16.4 m/s. its final vertical velocity, v is thus

V² = u² + 2gs where s = height of bridge above water = 16.4 m

v² = 16.4² + 2 * 9.8 * 16.4 = √ (268.96 + 321.44)

v = √590.4 = 24.30 m/s