ch04-p087 - 2 2 87. Using Eq. 2-16, we obtain v 2 = v0 2 gh...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
87. Using Eq. 2-16, we obtain 22 0 2 vv g h =− , or 0 () / 2 . hvv g (a) Since 0 v = at the maximum height of an upward motion, with 0 7.00 m/s v = , we have (7.00m/s) /2(9.80m/s ) 2.50 m. h == (b) The relative speed is 0 7.00 m/s 3.00 m/s 4.00 m/s rc vvv =−= = with respect to the floor. Using the above equation we obtain (4.00 m/s) / 2(9.80 m/s ) 0.82 m. h (c) The acceleration, or the rate of change of speed of the ball with respect to the ground
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online