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?
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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!
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