Asked by marco

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?
Answered by drwls
Vx = (175 m)/(time to fall)

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

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