A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand in the second trial 4H. Compare the FINAL VELOCITY'S for the packages right as they hit the ground (or right before)?

The final speed of the second package is twice as much as the final speed of the first package.

Explanation:

Free Fall Motion

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

And the distance traveled downwards is:

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

Replacing into the first equation:

Rationalizing:

Let's call v1 the final speed of the package dropped from a height H. Thus:

Let v2 be the final speed of the package dropped from a height 4H. Thus:

Taking out the square root of 4:

Dividing v2/v1 we can compare the final speeds:

Simplifying:

The final speed of the second package is twice as much as the final speed of the first package.

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

i guess it’s a and b