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Question

http://i80.photobucket.com/albums/j172/XwakeriderX/Cannon.jpg http://i80.photobucket.com/albums/j172/XwakeriderX/Cannon.jpg Just in case you cant read it Please go step by step As shown in the sketch above, a cannon, situated on level ground at the base of a tall hill with a slope angle of 30.0degree . fires a cannonball at an initial speed of 350.0 m/s at a 45.0degree angle from the horizontal The projectile flies toward the hill and impacts some where on the stope. Find the height above the ground at which the impact occurs. You may ignore air friction and the height of the mu zzle of the cannon above the ground.

Explanation / Answer

angle A = 45.0 degrees, angle B = 30.0 degrees, initial velocity v = 350.0 m/s find h horizontal displacement x = h/tan(B) = vcos(A)*t (1) vertical displacement y = h = vsin(A)*t - gt^2/2 (2) from (2), gt^2/2 - vsin(A)*t + h = 0 solve it and get t = {vsin(A) + sqrt[v^2*sin(A)^2 - 2gh]}/g put it into (1), get h/tan(B) = vcos(A)*{vsin(A) + sqrt[v^2*sin(A)^2 - 2gh]}/g gh/tan(B) = v^2*cos(A)*sin(A) + v*cos(A)*sqrt[v^2*sin(A)^2 - 2gh]} sqrt[v^2*sin(A)^2 - 2gh] = gh/[vcos(A)tan(B)] - v*sin(A) insert known values, simplify and get sqrt(61250 - 19.6h) = 0.0685857h - 247.487 squared both sides and simplify, 0.004704h^2 - 14.348h - 0.18483 = 0 solve it and get h = 3050 m

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