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Unformatted text preview: 100 m. A 5.00-m high fence is located 85.0 m horizontally form the launch point. a) What is the magnitude of the ball’s initial velocity? The equation for horizontal range yields R = (v 2 /g)sin2 θ ⇒ v = √ (Rg/sin2 θ ) = 33.64 m 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 θ = 29.13 m/s, v 0y = v sin θ = 16.82 m/s The ball reaches the position of the fence at Δ x = v 0x t t = Δ x/v 0x = 2.918 s At this moment, the height of the ball is y(t = 2.918 s) = y + v 0y t – (1/2)gt 2 = 7.86 m Therefore, it is 2.86 m (=7.86 – 5) above the fence....
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This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.
- Spring '08