Lesson_1.5_Printable

Lesson_1.5_Printable - Problem Solution Strategy for...

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Unformatted text preview: Problem Solution Strategy for projectiles: 1. Identify v o and 2. Obtain v x = v o cos and v yo = v o sin 3. The x-position and the y-position are: 2 1 os sin v t y v t gt =- ( 29 ( 29 cos sin 2 o o x v t y v t gt = =- 4. The horizontal and vertical velocity of the projectile at any instant are: v x = v x0 v y = v y0 gt 5. Use these information in the equations of motion to solve for the unknown. 1 An artillery shell is fired with an initial velocity of 100 m.s-1 at an angle of 30 above the horizontal. Find: a. Its position after 8 s v o = 100 m.s-1 , = 30 v x = 100 cos 30 x = ( v o cos ) t t = 8 s v yo = 100 sin 30 = 86.6 m.s-1 = 50 m.s-1 = 86.6(8) = 692.8 m Position is: 692.8 i + 86.4 j y = v yo t gt 2 = 86.4 m = 50(8) (9.8)(8) 2 2 An artillery shell is fired with an initial velocity of 100 m.s-1 at an angle of 30 above the horizontal. Find: b. Its velocity after 8 s v o = 100 m.s-1 , = 30 v x = 100 cos 30 t = 8 s v yo = 100 sin 30 = 86.6 m.s-1 = 50 m.s-1 v y = v yo gt 86.6 ms-1 = 50 (9.8)(8) = - 28.4 m.s-1 v x = 86.6 ms v y v 2 2 x y v v v = + 2 2 86.6 28.4 = + = 91.1 m.s-1 1 28.4 tan 36.6 -- = = -37.8 o 3 An artillery shell is fired with an initial velocity of 100 m.s-1 at an angle of 30 above the horizontal. Find: c. The time required to reach its maximum height. v o = 100 m.s-1 , = 30 v x = 100 cos 30 t = 8 s v yo = 100 sin 30 = 86.6 m.s-1 = 50 m.s-1 v y = v yo gt When the shell reaches its maximum height, v y = 0 yo v t g = = 5.1 s 0 = v yo gt-1-2 50 ms 9.8 ms = 4 An artillery shell is fired with an initial velocity of 100 m.s-1 at an angle of 30 above the horizontal. Find: c. The horizontal range of the shell....
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This note was uploaded on 05/01/2011 for the course PHY 2048 taught by Professor George during the Fall '10 term at Edison State College.

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Lesson_1.5_Printable - Problem Solution Strategy for...

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