CHAPTER 3 - 14 - 0 = ( v 2 sin ) 2 + 2( g ) h max , which...

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Unformatted text preview: 0 = ( v 2 sin ) 2 + 2( g ) h max , which gives h max = v 2 sin 2 /2 g . When we equate this to the range, we get v 2 sin 2 /2 g = 2 v 2 sin cos / g , which gives tan = 4, = 76.0 . 43. The ball passes the goal posts when it has traveled the horizontal distance of 36.0 m. From this we can find the time when it passes the goal posts: x = v x t ; 36.0 m = (20.0 m/s) cos 37.0 t , which gives t = 2.25 s. To see if the kick is successful, we must find the height of the ball at this time: y = y + v y t + ! a y t 2 = 0 + (20.0 m/s) sin 37.0 (2.25 s) + ! ( 9.80 m/s 2 )(2.25 s) 2 = 2.24 m. Thus the kick is unsuccessful because it passes 0.76 m below the bar....
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CHAPTER 3 - 14 - 0 = ( v 2 sin ) 2 + 2( g ) h max , which...

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