Answer:
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
