Using Kinematic Equation w/ Constant Velocity: Solving Time of Fall

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Homework Help Overview

The problem involves a scenario where a parachute and a stone are dropped from a height, with the parachute falling at a constant velocity. The participants are exploring the application of kinematic equations in this context, particularly focusing on the time it takes for each object to reach the ground.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations, particularly in scenarios involving constant velocity. There is an exploration of how to apply the equations when acceleration is zero, and questions arise about the correctness of the calculations and assumptions made.

Discussion Status

The discussion is ongoing, with some participants confirming the validity of the approaches taken. There is a sense of collaborative exploration, with guidance being offered regarding the simplification of equations when acceleration is zero.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available or the methods they can use. The specific conditions of the problem, such as the constant velocity of the parachute and the timing of the stone's drop, are central to the discussion.

celect
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How do I use Kinematic equation with constant velocity.
I only have used with intital and final velocity.



A parachute dropped from a 30-meter-high cliff falls with a constant
velocity of 1.5 meters per second. Twenty-two seconds later a stone is
dropped from the cliff.

a. How long does it take the parachute to hit the ground?


b. How long does it take the stone to hit the ground?
I solved this using d=v_i * t + 1/2 * a *t ^2

24.5 seconds
 
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Originally posted by celect
How do I use Kinematic equation with constant velocity.
I only have used with intital and final velocity.
You can use the exact same equation by setting a=0. It simplifies to d = v*t ... I hope that equation looks familiar!
 
d=v*t+1/2*a*[tex]t^2[/tex]

-30m = (-1.5m/s)*t + 0.5 (0m/s)*[tex]t^2[/tex]

I come up with t=20s

Am I going in the right direction?
 
yes, if the velocity remains contant
then a=0
 
Originally posted by celect
Am I going in the right direction?
Perfect.
 

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