4-2-09

.9143 = 4.9t ^2 .432 = t
 * = Vi ||= .9375 ||= 0 ||
 * = Vf ||= .9375 ||=  ||
 * = A ||= 0 ||= 9.8m/s ||
 * = x ||= .405m ||= .9143m ||
 * = t ||= .432 ||= .432 ||
 * = t ||= .432 ||= .432 ||

.405 = vi (t) .9375 m/s = vi

.9143 - .63 = .2813 m

.2813 m = 4.9t^2 .2396s = t

.9375 * .2396 = (x = .225m)

We did the Woolly projectile lab today. The main goal of the lab was to use projectile motion to decide where he should place the mamoth so that he could be sure to knock him out. We found the initial horizontal velocity by using th information we had avaialble to us like the height from the end of the ramp to the floor. . From there we used the information to find out the time. Once we found the time we used the time to find the intial horizontal velocity. We used the initial velocity to figure out the displacement of the ball which would be were we place the wholly mammonth according to whatever height we got for our test