You throw a beanbag in the air and catch it 2.2 s later at the same place at which you threw it. how high did it go? what was the initial velocity?

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)

answer: option c.

explanation:

pretty sure it's false