A cannon fires a 300.0lb cannonball at an angle of 45 degrees from the ground wi

Question

A cannon fires a 300.0lb cannonball at an angle of 45 degrees from the ground with an initial velocity of 350.0m/s. After 25.0 seconds, how much velocity in the y-direction remains? How much in the x-direction?

Explanation / Answer

the only force acting on the body is gravitational force which is acting vertically => there is no horizontal force so the horizontal velocity will remian constant => intial velocity in x-direction = velocity in x-direction after 25 sec = 350cos45= 247.5 m/s velocity in the y-direction after 25 sec is using v=u+at we get => v=u-gt =>v = (350*sin(45 degrees)) - (9.8*25) = 2.48 m/sec

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