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Unformatted text preview: 45. ( a ) We choose a coordinate system with the origin at the jump point, with x horizontal and y vertical, with the positive direction up. The horizontal motion will have constant velocity: x = x + v x t ; L = 0 + v cos θ t ; 3.0 m = ( v cos θ )(1.3 s), which gives v cos θ = 2.31 m/s. For the vertical motion we have y = y + v y t + ! a y t 2 ; – h = 0 + ( v sin θ ) t + ! (– g ) t 2 ; – 5.0 m = ( v sin θ )(1.3 s) – ! (9.80 m/s 2 )(1.3 s) 2 , which gives v sin θ = 2.52 m/s. When we divide the two equations, we get tan θ = 1.093, θ = 47.5°. Thus we find the magnitude of v from v sin 47.5° = 2.52 m/s, which gives v = 3.42 m/s, so v = 3.42 m/s, 47.5° above the horizontal . ( b ) At the maximum height the vertical velocity will be zero. We find the maximum height from) At the maximum height the vertical velocity will be zero....
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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 University of California, Berkeley.
- Fall '08