Unformatted text preview: 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 ) = 34.5 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 = 29.9 m/s, v 0y = v sin θ = 17.3 m/s The ball reaches the position of the fence at Δ x = v 0x t t = Δ x/v 0x = 3.18 s At that moment, the height of the ball is y(t = 3.18s) = y + v 0y t – (1/2)gt 2 = 6.66 m Therefore, it is 1.66 m (= 6.66 – 5) above the fence....
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 Summer '08
 Field
 Work, Velocity, Expression

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