Sam tosses a ball horizontally off a footbridge at 3.2 m/s. How much time passes after he releases it until its speed doubles?

Answer 1

The vertical velocity component Vy would have to increase to sqrt3 times the initial horizontal velocity, Vx, so that V^2 = Vx^2 + Vy^2 = 4 Vx^2, and
V = 2 Vx

To achive that velocity, it must fall a distance H such that

M g H = (1/2) M Vy^2 = 3/2 M Vx^2

H = [3/(2g)]*Vx^2

Answer 2

i don't understand your work. please try to explain it more thoroughly.