This preview shows page 1. Sign up to view the full content.
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...
View
Full
Document
This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.
 Summer '08
 Field
 Work

Click to edit the document details