Velocity of a freely falling object

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Homework Help Overview

The discussion revolves around a problem involving the motion of a freely falling object, specifically focusing on its velocity and distance traveled before hitting the ground. The object is released from rest and takes 1.65 seconds to cover the last 39.0 meters of its fall. Participants are exploring the implications of gravitational acceleration and the effects of air resistance on falling objects.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to find the velocity and distance of the falling object. There is an attempt to clarify the relationship between time, distance, and acceleration. Questions are raised about the dependence of falling velocity on mass and the role of aerodynamic factors.

Discussion Status

The discussion is ongoing with various interpretations being explored. Some participants have provided guidance on using specific equations, while others are questioning the assumptions related to free fall and the effects of air resistance. There is no explicit consensus on the interpretations of the concepts discussed.

Contextual Notes

Participants are navigating the constraints of the problem, including the requirement to indicate direction with the sign of the velocity and the implications of rounding in calculations. The discussion also touches on the definition of free fall and the conditions under which it applies.

ashleyFGCU
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Homework Statement


A certain freely falling object, released from rest, requires 1.65 s to travel the last 39.0 m before it hits the ground.
(a) Find the velocity of the object when it is 39.0 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.)
(b) Find the total distance the object travels during the fall.


Homework Equations


v=v°+at
a=v-v°/t
distance Δx=vavg(t)


The Attempt at a Solution


a) To find the velocity I am simply using the velocity formula, and making a=-9.8 because it is falling down.
v=0+1.65s(-9.8)=-16.17m
When I put -16.17 into the answer online it response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.
I also tried -16.2, but before I use another try (I only have 3 left) I wanted to make sure that I am close to the answer, so I can try rounding to 16m/s.
b) I am confused on this one but I got 47.1m and it tells me the same thing, that I am within 10% of the correct value.
 
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You have distance (-39.5 meters), you have time (1.65 seconds) and you have acceleration (-9.8 meters/s2). You should know the formula s = v0t + at2/2. So you should be able to find v0 from this equation. That's part 1.

For part 2, you have v0 that the object acquired after falling for some unknown time. For this, you should be able to find out the time. Then use the distance formula again to find out the total distance.
 
Does velocity of freely falling object depends on mass?
 
Arun Veera said:
Does velocity of freely falling object depends on mass?

No. Acceleration due to gravity is a constant g=9.807 m/s^2 for all objects regardless of mass. Only aerodynamic considerations will affect the velocities.
 
wreckemtech said:
No. Acceleration df2f
ue to gravity is a constant g=9.807 m/s^2 for all objects regardless of mass. Only aerodynamic considerations will affect the velocities.
what do mean by aerodynamic consoderations?! consider a book and a large car dropped from a certain height. car reaches first and That is due to the surface area of the car more then book and so atmosphEric pressure acting on the car is more , so car reaches ground first. Am I right? But the velocity that the both the bodies hit the ground is same.
 
Arun Veera said:
what do mean by aerodynamic consoderations?! consider a book and a large car dropped from a certain height. car reaches first and That is due to the surface area of the car more then book and so atmosphEric pressure acting on the car is more , so car reaches ground first. Am I right? But the velocity that the both the bodies hit the ground is same.

A larger surface area in relation to the mass of an object is likely to slow it. Think "wind resistance" slowing the acceleration of an object. In a vacuum, a feather and a bowling ball dropped from any height will hit the ground at the same time. Whereas in actual atmospheric conditions, the feather will fall much more slowly than the bowling ball because the feather has a huge surface area in relation to its mass.
 

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