# Maximum Height

Lucretius
A passenger in a helicopter traveling upwards at 28 m/s accidentally drops a package out the window. If it takes 19 seconds to reach the ground, how high to the nearest meter was the helicopter when the package was dropped?

The answer I got for this is: 1237 (to the nearest meter)

This is not my problem however, the next question is, and it reads:

To the nearest meter what was the maximum height of the package above the ground in the previous problem?

I thought maybe I would use the distance formula again, with x=v0(t)+1/2at^2

The number I get isn't right…the answer is 1279.

Gold Member
MHB
Well since the helicopter was traveling upwards, I would say the maximum height would be when the package was dropped. The helicopter was moving at a constant speed, meaning no acceleration. So once the package was released, the only force acting on it was gravity, making it head downward immediately.

That's my reasoning.

Lucretius
Remember that the package was traveling upwards at 28 m/s (this was initial velocity). So when the package is dropped, it still had some upwards motion. I need to figure out the distance covered from v=28m/s to v=0.

Gold Member
MHB
Ok. Sorry it's late. Use the kinematic equation you listed in your OP. Just make sure that you include direction, because the initial velocity and the acceleration are in different directions. It seems straightfoward from there.

Lucretius
What numbers would I be plugging in? If I plug in the 28, 19, and 9.8 I just get my first answer: 1237. 1237=28(19)-1/2(9.8)(19^2)

Homework Helper
Lucretius said:
What numbers would I be plugging in? If I plug in the 28, 19, and 9.8 I just get my first answer: 1237. 1237=28(19)-1/2(9.8)(19^2)
Nope. That is wrong. Why you choose t = 19s?
You know the plane is 1237 m above the ground, when the package is dropped.
The package will reach its maximum height when its velocity is 0. Because the package's initial velocity is 28 m / s. And it decelerates to 0 m / s. After that it will start to fall down. So when the package's velocity is 0, it reaches its max height.
How long does it take for the package to decelerate to 0 m / s?
Viet Dao,

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Lucretius
Oh, I get it now!

First I use v=v0+at, and find T=~2.9

then I reuse x=28(2.9)-1/2(9.8)(2.9^2). X=39.99999, aka 40. So I get 1277. I guess an error of 2m is acceptable.