http://i80.photobucket.com/albums/j172/XwakeriderX/Cannon.jpg As shown in the sk

Question

http://i80.photobucket.com/albums/j172/XwakeriderX/Cannon.jpg 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 stop. 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

initial velocity of the cannon u = 350 m/s angle of projection ? =45^o height at which the impact occus is the maximum height of the projectile                 H = u ^2 sin ^2 ? / 2 g                     = ( 350 ^2 ) ( sin 45 )^2 / 2 *9.8                     =    3125 m         

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