A cannon is fired horizontally from atop a 40.0 m tower. The cannonball travels 175 m horizontally before it strikes the ground. With what velocity did the ball leave the muzzle?

Answer 1

Vx = (175 m)/(time to fall)

Time to fall = sqrt(2 H/g) = 2.86 s

Solve for Vx

Answer 2

how did u calculate "time to fall". and what is H and g in the equation:
sqrt(2 H/g) = 2.86 s?