A cannon is fired horizontally from atop a 40.0 m tower. The cannonball travels 145 m horizontally before it strikes the ground. With what velocity did the ball leave the muzzle?

If you are "marco", you clearly need a tutor full time. H means height, g is a very universal symbol for the acceleration due to gravity.

http://www.jiskha.com/search/index.cgi?query=A+cannon+is+fired+horizontally+from+atop+a+40.0+m+tower.+The+cannonball+travels+145+m+horizontally+before+it+strikes+the+ground.+With+what+velocity+did+the+ball+leave+the+muzzle%3F+

This looks like a question I answered last week.

http://www.jiskha.com/display.cgi?id=1292888446

A cannon is fired horizontally from atop a 40.0 m tower. The cannonball travels 145 m horizontally before it strikes the ground. With what velocity did the ball leave the muzzle?

To determine the velocity at which the cannonball left the muzzle, we can use the principle of conservation of energy. We know that the gravitational potential energy at the top of the tower is converted into kinetic energy as the cannonball travels horizontally.

The gravitational potential energy (PE) at the top of the tower can be calculated using the formula: PE = m * g * h, where m is the mass of the cannonball, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the tower (40.0 m).

The kinetic energy (KE) of the cannonball as it hits the ground can be calculated using the formula: KE = (1/2) * m * v^2, where v is the velocity of the cannonball.

Since the cannonball is fired horizontally, there is no change in the vertical component of its velocity. Therefore, the change in gravitational potential energy is equal to the change in kinetic energy:

m * g * h = (1/2) * m * v^2

Canceling out the mass (m) on both sides of the equation, we get:

g * h = (1/2) * v^2

Now we can solve for the velocity (v):

v^2 = 2 * g * h

v = √(2 * g * h)

Substituting the values in:

v = √(2 * 9.8 m/s^2 * 40.0 m)

v = √(784)

v ≈ 28 m/s

Therefore, the cannonball left the muzzle at a velocity of approximately 28 m/s.