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Projectile Motion. Do not try this at home!. Projectile Motion. Figure 4-16 shows a pirate ship, moored 560 m from a fort defending the harbor entrance of an island. The harbor defense cannon, located at sea level, has a muzzle velocity of 82m/s.

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## Projectile Motion

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**Projectile Motion**Do not try this at home!**Projectile Motion**• Figure 4-16 shows a pirate ship, moored 560 m from a fort defending the harbor entrance of an island. The harbor defense cannon, located at sea level, has a muzzle velocity of 82m/s. (a) To what angle must the cannon be elevated to hit the pirate ship? (b)What are the times of flight for the two elevation angles calculated above? (c) How far the pirate ship be from the fort if it is to be beyond range of the cannon?**Equations of motion along a straight line with constant**acceleration**Projectile Motion**The range R is the horizontal distance the Projectile has traveled when it returns to its launch height The horizontal motion and the vertical motion are independent of each other.**The Horizontal Motion**• Because there is no acceleration in the horizontal direction, the horizon component of the projectile’s initial velocity remains unchanged throughout the motion, as demonstrated in the following figure:**Home work**• A movie stuntman is to run across a rooftop and jump horizontally off it, to land on the roof of the next building . Before he attempts the jump, he wisely asks you to determine whether it is possible. Can he make the jump if his maximum rooftop speed is 4.5m/s?**Home work**• A rescue plane is flying at a constant elevation of 1200 m with a speed of 430km/h toward a point directly over a person struggling in the water ( see Fig.4-14). At what angle of sight should be pilot release a rescue capsule is it is to strike (very close to) the person in the water?

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