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

**Unformatted text preview: **47. We neglect air resistance, which justifies setting a = -g = -9.8 m/s2 (taking down as the -y direction) for the duration of the motion. We are allowed to use Table 2-1 (with y replacing x) because this is constant acceleration motion. The ground level is taken to correspond to the origin of the y axis. (a) Using y = v0 t - 1 gt2 , with y = 0.544 m and t = 0.200 s, we find 2 v0 = y + 1 gt2 0.544 + 1 (9.8)(0.200)2 2 2 = = 3.70 m/s . t 0.200 (b) The velocity at y = 0.544 m is v = v0 - gt = 3.70 - (9.8)(0.200) = 1.74 m/s .
2 (c) Using v 2 = v0 - 2gy (with different values for y and v than before), we solve for the value of y corresponding to maximum height (where v = 0). 2 v0 3.72 = = 0.698 m . 2g 2(9.8) y= Thus, the armadillo goes 0.698 - 0.544 = 0.154 m higher. ...

View Full
Document