PHYS_2014_Lecture_5

# PHYS_2014_Lecture_5 - Fall 2010 Oklahoma State University...

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Unformatted text preview: Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 5, Slide 1 Leture 5: Projectile Motion (or shooting stuff) θ a v a R a h If I shoot an arrow into the air with initial velocity and at angle θ , it follows a parabolic trajectory. The horizontal ( x ) distance it travels is called the range, . The maximum vertical ( y ) distance it reaches half way through its flight we will call the height, . a v a R a h Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 5, Slide 2 As in the inclined track example, we must treat the x and y components of motion separately (two 1-D problems instead of one 2-D problem). The initial velocity in the x direction will be The initial velocity in the y direction will be cos = a a x v v θ sin = a a y v v θ θ sin = a a y v v θ cos = a a x v v θ a R a h Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 5, Slide 3 When the arrow reaches its maximum height, , its vertical velocity, , will be zero. At this point, by eq. 1: a y v 0m s y y v v gt = + = a a a sin v t g θ = θ m sin s = = a a y v v θ a R a h a h Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 5, Slide 4 The arrow takes the same time to fly from its maximum height back to the ground as it takes to reach its maximum height from the ground. The total time of flight is then: 10 20 30 40 50 60 70 80 90 100 Range (m) 5 10 15 20 25 Height (m) 2 sin 2 = = v T t g θ 1 2 3 4 5 Time (s) 5 10 15 20 25 Height (m) Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 5, Slide 5 Since there is no acceleration in the x direction, the horizontal velocity of the arrow is constant. cos = a a x v v θ The range, , then will just be the time of flight, T , times the horizontal velocity, . ( ) 0 0 2 2 sin 2 cos sin cos ˆ 2 sin cos x v v v R v T v g g v i R g θ θ θ θ θ θ = = = = a a a a a Apply the trignometric identity 1 sin cos sin 2 2 θ θ = θ and the total range will be: a R a x v 2 ˆ sin 2 = a o v i R g θ Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 5, Slide 6 What angle leads to a projectile traveling its maximum range? 2 ˆ sin 2 = a v i R g θ This is the same as asking, for what angle θ is sin2 θ a maximum? 90 180 270 360 Angle (degrees)-1.1-0.9-0.7-0.5-0.3-0.1 0.1 0.3 0.5 0.7 0.9 1.1 Maximum angle is 45 ° , since 2 × 45 ° = 90 ° . Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 5, Slide 7 What angle leads to the maximum height? Height will be maximum when all of is in the vertical ( y ) direction and there is minimum (zero) horizontal ( x ) component. Not surprisingly, the maximum height is reached by shooting the arrow straight up. a v Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 5, Slide 8 10 20 30 40 50 60 70 80 90 100 Range (m) 5 10 15 20 25 30 35 Height (m) 30 deg....
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PHYS_2014_Lecture_5 - Fall 2010 Oklahoma State University...

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