Projectile Motion Do not try this at home!.

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Introduction to 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

(Cont’d)

Projectile Motion The horizontal motion and the vertical 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:

The Horizontal Motion

The Vertical Motion

The Vertical Motion

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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?