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Lesson 1.5 Relative Motion and Problem Solving

# Lesson 1.5 Relative Motion and Problem Solving - Problem...

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Problem Solution Strategy for projectiles: 1. Identify v o and θ 2. Obtain v x = v o cos θ and v yo = v o sin θ 3. The x-position and the y-position are: ( 29 ( 29 2 1 cos sin 2 o o x v t y v t gt θ θ = = - 4. The horizontal and vertical velocity of the projectile at any instant are: 5. Use these information in the equations of motion to solve for the unknown. 1

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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 Position is: 692.8 i + 86.4 j t = 8 s = 50 m.s -1 = 86.6(8) = 692.8 m = 86.4 m 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 t = 8 s = 86.6 m.s -1 = 50 m.s -1 = 50 – (9.8)(8) = - 28.4 m.s -1 v 2 2 x y v v v = + 2 2 86.6 28.4 = + θ 1 28.4 tan 36.6 θ - - = = -37.8 o 3

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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. t = 8 s = 86.6 m.s -1 = 50 m.s -1 yo v t g = = 5.1 s -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. t = 8 s = 86.6 m.s -1 = 50 m.s -1 If the time taken to reach the maximum height is 5.1 s, the time of flight = 2 × 5.1 = 10.2 s = 883.7 m 5

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A plastic ball that is projected with a velocity of 15 m.s -1 stays in the air for 2.0 s.
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