<p>So I'm looking at this problem, and I just can't figure it out..</p>
<p>A cannon with a muzzle speed of 1000 m/s is used to start an avalanche on a mountain slope. The target is 2000 m from the cannon horizontally and 800 m above the cannon. At what angle, about the horizontal, should the cannon be fired?</p>
<p>Maybe it's a trig thing I'm missing.. but I found this problem to be a pain in the arse.. arghhh!</p>
<p>Use tangent...arctan x = 800/2000.</p>
<p>There is such a thing called a "Range Formula". It is derived in my textbook, but it is very difficult to type it. Sorry.</p>
<p>Well, you can do it from scratch... :)</p>
<p>If the gun is tilted at y degrees from the horizontal, and the round takes t seconds to hit the target, then</p>
<p>S = (init.vertical velocity)t - 0.5(9.8)t^2
or (1000 sin(y))t - 0.5(9.8)t^2 = 800 ----Eqn(1)</p>
<p>and (horizontal velocity)t = 2000
or (1000 cos(y))t = 2000 ----Eqn(2)</p>
<p>Simplify eqn(2) to get cos(y) = 2/t
Simplify eqn(1) to get sin(y) = (0.8/t) + (4.9/1000)t</p>
<p>Use sin^2(y) + cos^2(y) = 1 and substitute z = t^2 to get a quadratic equation in z.
Solve for z, work backwards to get t and then angle y.</p>
<p>whups, I made a mistake.</p>
<p>It should be tan x = 800/2000
So x = arctan (8/20)
x = arctan (2/5)</p>
<p>Time shouldn't be a factor in this problem, so the velocity is unneccessary, I think.</p>
<p>Actually I think optimizerdad got the right answer, because somebodynew you're assuming that the cannonball will travel in a perfectly straight line towards the target... which it won't. As it is, the cannonball accelerates downwards due to gravity so you have to compensate by making the angle fired from slightly greater than arctan(2/5). That's where the velocity (and optimizerdad's calculations) come in.</p>
<p>I had to do the same problem for hw problems two weeks ago. Range equation gets the right answer...I think you use it to find the angle and then put the angle into y equation (bullet is dropping so you know g, theta, etc). Find the appropriate equation and plug in. </p>
<p>You must have the new book. My book is old and has everything in feet...sigh, when I started problem I kept getting the wrong answer, and then I looked at units</p>
<p>Oh yeah, whups...then yeah, the range equation will get you the right answer.</p>
<p>
When you start scratching, it's hard to stop! ;)
g=9.8
Free quote from optimizerdad:
Height H = (init.vertical velocity)t - (g/2)t^2
or 800 = (1000 sin(y))t - (g/2)t^2 ----Eqn(1)</p>
<p>and distance D = (horizontal velocity)t
or 2000 = (1000 cos(y))t ----Eqn(2)
End of quote.</p>
<p>From Eqn(2)
t = 2000/(1000 cos(y)) = 2/cos(y).
Plugging into Eqn(1)
800 = (1000 sin(y)) 2/cos(y) - (g/2)(2/cos(y))^2
800 = 2000 tan(y) - 2g/(cos(y))^2</p>
<p>Identity 1/(cos(y))^2 = (tan(y))^2 + 1 to the resque!
Proof:
(tan(y))^2 + 1 =
(sin(y))^2 / (cos(y))^2 + 1 =
((sin(y))^2 + (cos(y))^2) / (cos(y))^2 =
1/(cos(y))^2.</p>
<p>800 = 2000 tan(y) - 2g (tan(y))^2 + 1)
Substitute z = tan(y)
2gz^2 - 2000z + 800 + 2g = 0
2 9.8 z^2 - 2000z + 800 + 19.6 = 0
19.6 z^2 - 2000z + 819.6 = 0
z1 = .41145912640446
z2 = 101.62935720013</p>
<p>Back to tan(y) = z
tan(y1) = .41145912640446
tan(y2) = 101.62935720013</p>
<p>y1 = 22.365163229244 deg
y2 = 89.436246257108 deg</p>
<p>These are the two possible angles at which cannon can be fired to start an avalanche.</p>
<p>I am not sure I've got enough loose nuts to stand next to that loose cannon when it fires almost straight up in the air (at 89 deg) and wait to see who gets hit: me or an avalancheous mountain slope. :eek:</p>
<p>Hard to believe that 89+ degree value at first glance, but it makes sense. With a muzzle velocity of 1000 m/s and a vertical component only slightly less, the shell would take slightly less than (1000/9.8) = 102 seconds to reach its peak, plus almost all this time to fall back to a height of 800 m. That gives us a hang time of around 200 seconds!</p>
<p>Thanks for the responses!</p>
<p>I had the concept down, but I couldn't do the trig.. it was a beast! We went over it during class, and my teacher told us to forget about the trig (it was the only problem that had such complex trig identities!) and just get the concept down. I appreciate the input though!!</p>
<p>can you not use the equation</p>
<p>R= (V0^2 - 2sin(theta))/(g) </p>
<p>where R is the horizontal distance between target and cannon?</p>
<p>Thanks</p>
<p>I can’t solve this problem, but I have the answer: 22,4 deg or 89,4 deg</p>
<p>Thanks for the confirmation!</p>