Acannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. if the cannon is on a mounted wagon 1.5 m tall, how far would the cannonball travel before it lands on the ground? (show all work)

473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

  (1)

Where:

is the cannonball's initial velocity

because we are told the cannonball is shot horizontally

is the time since the cannonball is shot until it hits the ground

y-component:

  (2)

Where:

 is the initial height of the cannonball

 is the final height of the cannonball (when it finally hits the ground)

 is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it and this occurs when .

So, firstly we will find the time with (2):

  (3)

Rearranging the equation:

  (4)

  (5)

This is a quadratic equation (also called equation of the second degree) of the form , which can be solved with the following formula:

(6)

Where:

Substituting the known values:

(7)

Solving (7) we find the positive result is:

(8)

Substituting this value in (1):

  (9)

 This is the horizontal distance the cannonball traveled before it landed on the ground.

e) none of the above answers are correct

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6