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)

a. The object can have zero acceleration and, simultaneously, nonzero velocity.

c. The object can have nonzero velocity and nonzero acceleration simultaneously.

d. The object can have zero velocity and, simultaneously, nonzero acceleration.

Explanation:

For an object in simple harmonic motion, the total mechanical energy (sum of elastic potential energy and kinetic energy) is constant:

(1)

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

We can also notice that the force on the spring is given by Hook's law:

And since according to Newton's law we have F = ma, this can be rewritten as

which means that the acceleration is proportional to the displacement.

So by looking again at eq.(1), we can now states that:

- when the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum

- when the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

- in all the other intermediate situations, both velocity and acceleration are non-zero

So the correct answers are

a. The object can have zero acceleration and, simultaneously, nonzero velocity.

c. The object can have nonzero velocity and nonzero acceleration simultaneously.

d. The object can have zero velocity and, simultaneously, nonzero acceleration.