Kinetic projectile motion (Helicopter questions)

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Homework Help Overview

The problem involves a helicopter flying upward while a coke bottle is dropped from it. The objective is to determine the time it takes for the bottle to hit the ground, given the initial height and the upward velocity of the helicopter. The context is within the subject area of kinematics, specifically projectile motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various equations of motion, with some suggesting to avoid certain calculations and methods like factoring or using quadratics. There is a focus on finding a suitable formula that adheres to the homework constraints.

Discussion Status

The discussion is ongoing, with participants offering different approaches and questioning the restrictions on methods. Some guidance has been provided regarding the use of specific equations, but there is no clear consensus on how to proceed without factoring.

Contextual Notes

There are constraints mentioned regarding the prohibition of factoring or using quadratic equations, which some participants express confusion about. This restriction is said to be in place to prepare students for future coursework.

bbalin23
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Homework Statement



A helicopter flies upward at 3.0m/s. The pilot drops a coke bottle when the helicopter is 30.0m above ground. How long will it take the coke bottle to hit the ground?
(2.8s)

*Note* We can't factor or use quadratics to solve

Homework Equations


The Attempt at a Solution



First I tried to calculate the distance the coke bottle will go upward before falling downward:

Vf2-Vi2/2a = d

9/19.62

0.459

Then I added that to the distance... D(total) = 0.459m + 30.0m = 30.459m

Then I attempted to find the time from there using the formula:

d = vi(t) + 1/2 a t(squared)

Vi from the maximum height (Total distance) is zero so

Sqrt(2d/a) = t

Sqrt(60.918/9.81)
Sqrt(6.20978)
t = 2.4919
= 2.5s

Im off by .3 please help!
 
Last edited:
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Just use the equation normally. Don't calculate the first thing you did.

So basically just use:
d=v_{i}t-4.9t^{2}+h
h being the initial height.
Then you'll end up factoring and have 1 of the answers
 
iRaid said:
Just use the equation normally. Don't calculate the first thing you did.

So basically just use:
d=v_{i}t-4.9t^{2}+h
h being the initial height.
Then you'll end up factoring and have 1 of the answers

Sorry I forgot to mention it earlier but we weren't supposed to factor or use quadratic equations :(
 
You only need one formula,

y = y_o + v_o*t + a*t^2/2

You know y_o, v_o, and a
 
Spinnor said:
You only need one formula,

y = y_o + v_o*t + a*t^2/2

You know y_o, v_o, and a

I'm not sure which formula that is unfortunately as its not on my formula sheet, so I don't think I would be allowed to use it.
 
Why can't you factor lol?
 
iRaid said:
Why can't you factor lol?

Apparantly we are supposed to be able to figure this out without factoring (She says its to refrain us from using calculus later on in the course as if we do we won't get marks on the question)
 
Thanks everyone I figured it out using a couple different formulas! Thanks for helping!
 

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