If the total trip took
2.2 secs, then it must took 1.1 secs to reach max height.
We can then find the max height by realizing that the bag
fell from rest and took 1.1 secs to come back to its launch point, we use formula:
hmax = 1/2 gt^2 = 1/2
(9.8m/s/s)(1.1s)^2
hmax =5.93m (ANSWER)
To get the initial speed, we take that the vertical speed is
zero at max height, and use
vf^2=v0^2+2ad
vf=final speed =0
v0=intiial speed
a=acceleration = -9.8m/s/s
d=distance traveled =hmax= 5.93m
0=v0^2+2(-9.8ms/s/)(5.93m)
v0^2=2x9.8x5.93
v0=10.8m/s (ANSWER)
