Question

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)?

Answer 1

The final velocity of a package dropped from an air balloon depends on the time taken to fall and the acceleration due to gravity. The formula to calculate the final velocity is v = gt. By finding the time of fall in both trials and substituting it into the formula, we can compare the final velocities of the packages.

Step-by-step explanation:

When a package is dropped from an air balloon, it falls under the influence of gravity. In this case, we can assume that there is no air resistance. The final velocity of an object in free fall is determined by the time it takes to fall and the acceleration due to gravity. The formula to calculate the final velocity is v = gt, where v is the final velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth), and t is the time of fall.

In the first trial, the package is dropped from a height H. The time it takes to fall can be calculated using the equation h = (1/2)gt^2, where h is the distance fallen. Rearranging the equation, we get t = sqrt(2h/g). Substituting the value of h = H, we find the time taken to fall in the first trial.

In the second trial, the package is dropped from a height 4H. Using the same equation, we can find the time taken to fall in the second trial. Once we have the time taken to fall in both trials, we can substitute it back into the equation v = gt to find the final velocities of the packages right before they hit the ground.

Answer 2

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

Step-by-step 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:

`v=gt`

And the distance traveled downwards is:

`\displaystyle y=(gt^2)/(2)`

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:

`\displaystyle t=\sqrt{(2y)/(g)}`

Replacing into the first equation:

`\displaystyle v=g\sqrt{(2y)/(g)}`

Rationalizing:

`\displaystyle v=√(2gy)`

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

`\displaystyle v_1=√(2gH)`

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

`\displaystyle v_2=√(2g(4H))`

Taking out the square root of 4:

`\displaystyle v_2=2√(2gH)`

Dividing v2/v1 we can compare the final speeds:

`\displaystyle v_2/v_1=(2√(2gH))/(√(2gH))`

Simplifying:

`\displaystyle v_2/v_1=2`

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