(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A cannon with a muzzle speed of 992 m/s is used to start an avalanche on a mountain slope. The target is 1900 m from the cannon horizontally and 793 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)

2. Relevant equations

Xf= xi+vit

Yf=yi+vit+1/2at2

vfy^2=viy^2+2a(yf-yi)

x=cos(theta)992m/s

Y=sin(theta)992m/s

3. The attempt at a solution

ok this one i can get all the equations setup.

xf= cos(theta)992m/s

793=(sin(theta)992m/s)t + 1/2(-9.80)t2

using the x eqaution i can solve for t.

t=1900/cos(theata)992m/s

but once i place it into the eqaution, it gets very messy and i get stuck on the arithmatic.

793= (sin(theta)992m/s)(1900/cos(theata)992m/s)+1/2(-9.80)(1900/cos(theata)992)^2

this is where i am stuck.

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# Homework Help: Projectle problem

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