15.0 I’m pretty sure that’s the answer to your question
Answer:
a. 
b. 
Explanation:
The inertia can be find using
a.





now to find the torsion constant can use knowing the period of the balance
b.
T=0.5 s

Solve to K'


Answer:

Explanation:
From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

Solving N from (2) and replacing in (1):

The centripetal acceleration is given by:

Replacing and solving for v:

A, hope this helped! I didn’t really get it but I think it’s correct?
Answer:
i don't know if this is good for you but
Explanation:
ignoring frictional air resistance (drag) the speed on return is the same as when it left the ground (5 m/s but in the opposite direction).
Note: this points out a good reason for not firing live bullets into the air..they will return somewhere and at the same speed.
However, if you take into account the atmospheric drag the reurn speed will be somewhat smaller (but in the case of a bullet, probably still lethal.) Drag depends on many factors and is difficult to calculate.