ch04-p125

# ch04-p125 - 125. At maximum height, the y-component of a...

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

v 0 y gt again but now “starting the clock” at the highest point so that v 0 y = 0 (and 1.0 s t = ). This leads to v y = –9.8 m/s and () 2 2 (10 m/s) 9.8 m/s 14 m/s +− = . (c) The x 0 value may be obtained from x = 0 = x 0 + (10 m/s)(1.0s), which yields 0 10m. x =− (d) With v 0 y = 9.8 m/s denoting the y -component of velocity one second before the top of the trajectory, then we have yy v t g t y == + 0 00 1 2 2 where t = 1.0 s. This yields 0 4.9 m. y (e) By using x x 0 = (10 m/s)(1.0 s) where x 0 = 0, we obtain x = 10 m. (f) Let t = 0 at the top with 0 y yv == . From 2 1 2 y y t g t −= , we have, for t = 1.0 s, 22 (9.8 m/s )(1.0 s) / 2 4.9 m. y 125. At maximum height, the y -component of a projectile’s velocity vanishes, so the
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