CHAPTER 3 - 12 - 20 30 40 50 10 20 40 60 80 100 y (m) x (m)...

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Unformatted text preview: 20 30 40 50 10 20 40 60 80 100 y (m) x (m) 15° 30° 90° 60° 45° 75° We find the time to fall from the highest point from y = y + v y t + ! a y t down 2 ; 0 = h + 0 + ! (– g ) t down 2 , which gives t down = (2 h / g ) 1/2 = [2( v y 2 /2 g )/ g ] 1/2 = v y / g , which is the same as t up . 35. To plot the trajectory, we need a relationship between x and y , which can be obtained by eliminating t from the equations for the two components of the motion: x = v x t = v (cos θ ) t ; y = y + v y t + ! a y t 2 = 0 + v (sin θ ) t + ! (– g ) t 2 . The relationship is y = (tan θ ) x – ! g ( x / v cos θ ) 2 . 36. The arrow will hit the apple at the same elevation, so we can use the expression for the horizontal range: R = v 2 sin(2 θ )/ g ; 25.0 m = (22.5 m/s) 2 sin(2 θ )/(9.80 m/s)/(9....
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This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at Berkeley.

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CHAPTER 3 - 12 - 20 30 40 50 10 20 40 60 80 100 y (m) x (m)...

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