Question

Sam tosses a ball horizontally off a footbridge at 3.1 m/s. How much time passes after he releases it until its speed doubles?

Answer 1

The horizontal projection is constant while the vertical is changing according to the equation
V_y=V_0y+gt
(we chose the downward direction as positive).
V_x = 3.1 m/s
V_0y = 0;
V₀ = 3.1 m/s
V = √(V_x² + V_y²) = √(3.1² +100t²)
V²/V₀² = 4
(3.1² +100t²)/ 3.1² = 4
t =√(3.1²*3 / 100) = 0.54 s