# Free Falling Objects: Find Time for Mailbag to Reach Ground

• pberardi
In summary, the height of a helicopter above the ground is given by h = 3.00t^3, where h is in meters and t is in seconds. After 2.00s the helicopter releases a small mailbag, and it takes approximately 7.96 seconds for the mailbag to reach the ground. This is determined by using the equation Xf = Xi + VxT + .5at^2 and taking the derivative of the height function to find the velocity function.
pberardi

## Homework Statement

The height of a helicopter above the ground is given by h = 3.00t^2, where h is in meters and t is in seconds. After 2.00s the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground

## Homework Equations

Xf = Xi + VxT + .5at^2

## The Attempt at a Solution

h = 24
a = -9.8

0 = 24 + 24t + .5(-9.8)t^2
t = 5.7

pberardi said:

## Homework Statement

The height of a helicopter above the ground is given by h = 3.00t^2, where h is in meters and t is in seconds. After 2.00s the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground

## Homework Equations

Xf = Xi + VxT + .5at^2

## The Attempt at a Solution

h = 24
a = -9.8

0 = 24 + 24t + .5(-9.8)t^2
t = 5.7

Wouldn't the initial height be 12? 2^2*3 = 12?

Also, do you do physics with calculus? If so, you could take the derivative of the height function to get the velocity function, and simply plug in for t. If not, you can only get the average velocity, which won't do you any good in this problem, since the helicopter is accelerating.

I am sorry, my mistake. I did do the calculus. But I wrote the height function incorrectly. It is supposed to be h = 3t^3

So based on this new information, is it correct?

pberardi said:
I am sorry, my mistake. I did do the calculus. But I wrote the height function incorrectly. It is supposed to be h = 3t^3

So based on this new information, is it correct?
The initial height is correct, but the derivative of 3t^3 is 9t^2. With that in mind, the initial velocity would be 36.

Should be
0 = 24 + 36t + .5(-9.8)t^2
t = 7.96
?

pberardi said:
Should be
0 = 24 + 36t + .5(-9.8)t^2
t = 7.96
?
That looks good.

Thank you sir.

They might deduct a couple points if units aren't shown in every step ...
0 = 24.0 m + (36.0 m/s) t + (1/2) (-9.8 m/s^2) t^2

Last edited:

## 1. How is the time for a free falling object to reach the ground calculated?

The time for a free falling object to reach the ground can be calculated using the formula t = √(2h/g), where t is the time in seconds, h is the height in meters, and g is the acceleration due to gravity (9.8 m/s²).

## 2. Does the mass of the object affect the time it takes to reach the ground?

No, the mass of the object does not affect the time it takes to reach the ground. The time is solely determined by the height and the acceleration due to gravity.

## 3. How does air resistance impact the time for a free falling object to reach the ground?

Air resistance does have an impact on the time for a free falling object to reach the ground. The more air resistance there is, the longer it will take for the object to reach the ground. However, this impact is typically negligible for most objects.

## 4. Can the time for a free falling object to reach the ground be different on different planets?

Yes, the time for a free falling object to reach the ground can be different on different planets. This is because the acceleration due to gravity varies on different planets, so the formula for calculating time will be different.

## 5. What is the difference between free fall and free fall acceleration?

Free fall refers to the motion of an object falling under the influence of gravity alone, with no other forces acting on it. Free fall acceleration, on the other hand, is the rate at which an object's velocity increases as it falls due to the acceleration of gravity. It is a constant value of 9.8 m/s² on Earth.

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