A cannon ball leaves a fort with an initial horizontal speed of 1.8*10^2 m/s and strikes a ship in the sea below 7.2 s later. How high is the fort above sea level?
How high is the fort?
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Initial height : Δy=½gt² =½(9.8)(7.2)² = 250 m.
Note that the answer is independent of the initial horizontal velocity.
Hey Sun -- constant g ... use your delta V's of 10 @ sec ... Vdn at sea is 72m/s ...
Vave is 36m/s ... freefall 36m/s for 7.2s ==> 210 +42 +7 ~260m ... Regards :)
This takes a little longer to explain, instead of looking up a strange formula on a
formula sheet or even worst trying to memorize the formulas - but I think it is easier.
You can find the answer using the same smarts you use to figure out how much money
you need to buy 10 candy bars a day for the next 7 days if candy bars are $10.00 (or $9.81).
You can use the formulas or think of it this way.
1 - One important thing to remember with falling object or X & Y problems
is that the X and Y are independent. Always.
So for this we Only need to solve a Y or falling problem.
2- What is the starting velocity? Zero - 0
3- After 7.2 seconds falling (when it hit the water) how fast was it going?
7.2 sec. X 10 meter per sec. increase every sec. = 7.2 s X 10 m/s2 = 72 m/s
(if you want to make the problem harder and be a minuscule
2.5% more accurate you can use 9.81 m/s2 instead of 10 m/s2)
4- Okay now you know the starting speed and the ending speed;
what was the average speed? (0 + 72) ÷ 2 = 36 so an average of 36 m/s
5- Good! Now, if something goes an average of 36 m/s for 7.2 s, how far does it go?
36 m/s for 7.2 s = 36 m/s X 7.2 s = 260 m
6) So it fell 260 m therefor the fort was 260 m above sea level.
(Using 9.81 instead of 10 would give 254 - we were 2.4 % high)