A cannonball was shot upwards out of a cannon from the top of a castle. The initial height of the cannonball was 40 m above the ground.Two...

In this problem we have 2 variables, the height of the
cannonball and the time.  The time is independent (x) and the height is dependent (y).
Using the information in the problem we can determine two
points:

(0, 40)             This means at 0 minutes, the
cannonball is at 40 m

(3, 0)     This means that a 3
minutes, the height is 0 meters

When writing an equation,
we need to find the slope.  We can do this using the slope
formula:

(y2 -
y1)/(x2-x1)

Plug your values of x and y above into this
formula and you get:

(0 - 40)/(3-0) =
-40/3

Now you have the slope (-40/3) as well as 2 points.
 Now input what you are given into the point-slope formula: y - y1= m(x -
x1).

Remember that m = the slope and x1 and y1 = either of
your original points.  In this case, I will you (3, 0), but you could also use (40, 0)
as your point if you prefer.

y - y1
=m(x-x1)

y - 0 = -40/3(x - 3)
Substitute

y = -40/3x + 40
Distribute

Last, remember how we said x represented seconds
and y represented the height?  The original question asked that the height was a 2.5
seconds.  So we plug in 2.5 for x into our equation and solve for y as
follows:

y = -40/3(2.5) +
40

y = -33.333 + 40

y = 6.667
feet (approximately)

After 2.5 seconds, the cannonball is
approximately 6.667 feet off the ground.   Keep up the good work and keep
practicing!