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 ball’s initial velocity? The equation for horizontal range yields R = (v 2 /g)sin2 θ ⇒ v = √ (Rg/sin2 θ ) = 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 θ = 30.55 m/s, v 0y = v sin θ = 17.64 m/s The ball reaches the position of the fence at Δ x = v 0x t t = Δ 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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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