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Unformatted text preview: 110 m. A 6.00-m high fence is located 95.0 m horizontally form the launch point. a) What is the magnitude of the balls initial velocity? The equation for horizontal range yields R = (v 2 /g)sin2 v = (Rg/sin2 ) = 35.28 m/s b) At the fence, what is the distance between the fence top and the ball center? The x and y components of the initial velocity are v 0x = v cos = 30.55 m/s, v 0y = v sin = 17.64 m/s The ball reaches the position of the fence at x = v 0x t t = x/v 0x = 3.11 s At this moment, the height of the ball is y(t = 3.11 s) = y + v 0y t (1/2)gt 2 = 8.47 m Therefore, it is 2.47 m (= 8.47 6) above the fence....
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- Spring '08