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 seconds after being shot, the cannonball was again at the height of 40 m above the ground. Three seconds after being shot the cannonball hit the ground. What is the height of the cannonball 2.5 seconds after being shot?
1 Answer | Add Yours
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!
We’ve answered 319,180 questions. We can answer yours, too.Ask a question