Posted by **Tram** on Tuesday, November 12, 2013 at 2:02am.

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??

- physic--help --- -
**Steve**, Tuesday, November 12, 2013 at 4:29am
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.

- physic--help --- -
**Tram**, Tuesday, November 12, 2013 at 9:03am
Where is 4.9 come from?

- physic--help --- -
**Tram**, Tuesday, November 12, 2013 at 9:16am
I got it , thank you !

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