Answer:t=0.3253 s
Explanation:
Given
speed of balloon is 
speed of camera 
Initial separation between camera and balloon is 
Suppose after t sec of throw camera reach balloon then,
distance travel by balloon is


and distance travel by camera to reach balloon is


Now






There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .
(b)When passenger catches the camera time is 
velocity is given by



and position of camera is same as of balloon so
Position is 

Answer:
4m/s2
Explanation:
acceleration=v-u/t...v=12m/s,u=0m/s,t=3sec
a=12-0/3
=4m/s2
To find or discover by investigation?
Answer:
F in the definition of potential energy is the force exerted by the force field, e.g., gravity, spring force, etc. The potential energy U is equal to the work you must do against that force to move an object from the U=0 reference point to the position r.
Explanation: