A cannon ball shoots out from a cannon that is 28 inches long at 2700 ft s. What is the

acceleration and its time outside the cannon.

a = 1565217.39 ft / s ^ 2

t = 0.001725 seconds

Step-by-step explanation:

The first thing is to use the same system of units therefore we will pass the 28 inches to feet, like this:

28 in * (1 ft / 12 in) = 2.33 ft

Now yes, we can continue, we have the following data:

vi = 0

vf = 2700 ft / s

the equations in this case are as follows:

vf = vi + a * t

vf = a * t

rearranging for a

a = vf / t (1)

now with the position equation we know that:

x = vi * t + (a * t ^ 2) / 2

x = (a * t ^ 2) / 2 (2)

now replacing (1) in (2), we are left with:

x = (vf / t) * (t ^ 2) / 2

knowing that x would be 2.33 ft, which is when the cannonball exits the cannon.

2.33 = 2700 * t / 2

t = 2.33 * 2/2700 = 0.001725 seconds.

and now replace in (1)

a = vf / t = 2700 / 0.001725 = 1565217.39 ft / s ^ 2

good luck!

cutie~

f

step-by-step explanation:

f