Question

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

The vertical movement of the projectile is described by:

y = H - gt² / 2

When the cannonball is on the ground, y= 0 so:

0 = H - gt² /2

Solving for t:

$t = \sqrt{\dfrac{2H}{g}} = \sqrt{\dfrac{2(40\;m)}{9.8\;m/s^2}} = 2.8\;s$

The horizontally movement of the projectile is described by:

x = v₀t

Solving for v₀:

v₀ = x/t

v₀ = 175 m / 2.8 s

v₀ = 61.5 m/s