posted by Tram on .
one of the main battery of cannons protecting the entrance to San Francisco Bay were able t horizontally propel 7.3 kg cannon balls with an average speed of 26.2m/s. if the height of the cannon emplacement were 112m above San Francisco Bay,
a. how far from the cannon's muzzle (end of cannon where cannon ball exits) would the cannon ball land?
b. at what would the cannon need to be set so as to land half the distance?
My answer: a. 124m . b. 151 degrees
Are my answers correct??
how long does it take to fall 112m?
4.9t^2 = 112
t = 4.78 seconds
horizontal distance traveled is
4.78s * 26.2m/s = 125.26m
Rounding could account for the difference in our answers. I assume your 151° fits the equation, but it seems a bit odd to be firing backwards. Look for an acute angle that will give the same answer.
Where is 4.9 come from?
I got it , thank you !