Unformatted text preview: We take +y direction upward. The projection angle is θ = 3º. The x and y components of the initial velocity is v 0x = v cos θ = 39.9 m/s, v 0y = v sin θ = 2.09 m/s The ball reaches the position of the net when the horizontal displacement is d = 13 m. We get d = v 0x t ⇒ t = d/ v 0x = 0.326 s b) When the ball reaches the net, what is the distance between the center of the ball and the top of the net? When the ball reaches the location of the net, the height of the ball is y(t = 0.326 s) = y + v 0y t – (1/2)gt 2 = 1.30 m. The distance between the ball and the top of the net is 1.30 – 0.85 = 0.45 m...
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 Summer '08
 Field
 Work, Velocity, initial velocity, Tomoyuki Nakayama, Discussion Section7193 Quiz

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