# Time it takes for an object to hit the ground

## Homework Statement

The height of a helicopter above the ground is given by h = 2.55t^3, where h is in meters and t is in seconds. At t = 1.65 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

## Homework Equations

The above equation, and d = 1/2at (because initial velocity is zero)

## The Attempt at a Solution

2.55t^3
2.55(1.65)^3 = 11.45491875 m/s

11.45491875 = 1/2 (-9.8) (t^2)
22.9098375 = (-9.8) (t^2)
-2.3377 = t^2

You then get an imaginary number, but regardless, because time cannot be negative, I pretended like the answer was positive and took the square root.

I got uhhh 1.528966488, but when I plug it into the online homework, it's incorrect?

## Homework Statement

The height of a helicopter above the ground is given by h = 2.55t^3, where h is in meters and t is in seconds. At t = 1.65 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

## Homework Equations

The above equation, and d = 1/2at (because initial velocity is zero)

You say the height of the helicopter (which contains the mailbag) is a function of time. The initial velocity therefore cannot be zero.

rl.bhat
Homework Helper
Since the h is the function of the time, the helicopter is rising. So when the mail bag is released, it's initial velocity is not equal to zero. Take the derivative of h and find the velocity of the mail bat at the time of release. It is in the upward direction.

h(t) = 2.55t^3
It means the height changes with time.
You can find velocity v(t).
Initial velocity not zero.

NascentOxygen
Staff Emeritus
The downdraft from the rotor would give the mailbag extra downward impetus. Are we to ignore this? utkarshakash
Gold Member
Hey. You are doing a very serious mistake. Since you have taken the downward direction as +ve(you took downward displacement of the ball to be positive in your equation), you must also take g to be +ve i.e. +9.8m/s. Also the ball's initial velocity wrt helicopter is 0 and not wrt ground. Take the time derivative of h and find the velocity of helicopter at 1.65 s. Then apply the concept of relative velocity to obtain the ball's velocity at the instant it is released. I have given you enough hints. Now you can solve this. ;)

Okay, initial velocity cannot be zero because the helicopter is constantly rising, correct?

So you take the first derivative of your position-time equation, and plug in the time given (1.65 sec) and you get 20.827125 m/s.

11.45491875 = (20.827125)(t) + 1/2(9.8)t^2

gravity being positive because we regarded the height (displacement) as positive, right?

Then you'd solve the above equation which i'm not sure how to do lol.

rl.bhat
Homework Helper
In the freely falling body, usually we take the ground as the reference point, or attach the co-ordinate axis to the starting point.
The displacement is the difference between the final position and the initial position.
In this problem it is negative. g is always in the downward direction. The quantities in the same direction must have the same sign. So rewrite your equation and solve the quadratic to find the time.

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