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Basic Kinematics Question

by KingKai
Tags: basic, kinematics
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KingKai
#1
Nov4-12, 07:40 PM
P: 31
The equation


v22 = vo2 + 2aΔy


is used to determine the change in position, acceleration, or velocity of a particle under the influence of a quadratic force.


on earth, the acceleration value in this equation would, under normal circumstances, be
a= -9.8 m/s2.


so, this substitutes into

v22 = vo2 + -19.6Δy

In this

Δy = v22 - vo2 / -19.6

Lets now assume that v22 is 0, the crest of a projectiles motion.

Δy = 0 - vo2 / -19.6


If the objects initial v is facing upwards, then obviously the net Δy would be positive.



Now, keeping all conditions the same, lets assume that the acceleration is 9.8m/ss upwards

This would yield

Δy = 0 - vo2 / 19.6

Δy would thus be negative, and the object would be thrown downwards, accelerating upwards.
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Simon Bridge
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Nov4-12, 10:49 PM
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I'm sorry - was there a question in all that?
Simon Bridge
#3
Nov4-12, 11:15 PM
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I just want to go through your reasoning to make sure I follow:
Quote Quote by KingKai View Post
The equation

v22 = vo2 + 2aΔy

is used to determine the change in position, acceleration, or velocity of a particle under the influence of a quadratic force.
That would be a net unbalanced force which is constant.
on earth, the acceleration value in this equation would, under normal circumstances, be
a= -9.8 m/s2.
"normal circumstances" being, +y = "upwards", Δy<<REarth, and gravity is the only force acting on the object.
so, this substitutes into

v22 = vo2 + -19.6Δy
Your value has changed somehow [x2 of course] ... better to put g=9.8N/kg so a=-g so it does not matter what units you want to work in just yet... you get:$$v_2^2=v_0^2-2g\Delta y$$
In this

Δy = v22 - vo2 / -19.6
$$\Delta y = \frac{v_2^2-v_0^2}{-2g} = \frac{v_0^2-v_2^2}{2g}$$
Lets now assume that v22 is 0, the crest of a projectiles motion.

Δy = 0 - vo2 / -19.6
$$\Delta y = \frac{v_0^2}{2g} > 0$$
If the objects initial v is facing upwards, then obviously the net Δy would be positive.
No problems there.
Now, keeping all conditions the same, lets assume that the acceleration is 9.8m/ss upwards
All conditions?
So you put gravity upwards, and the final velocity is zero, but the assumption that the initial velocity is also upwards is incorrect - as you found below:
This would yield

Δy = 0 - vo2 / 19.6
$$\Delta y = \frac{0-v_0^2}{2g}=-\frac{v_0^2}{2g} < 0$$
Δy would thus be negative, and the object would be thrown downwards, accelerating upwards.
Well ... you did specify that the final velocity is zero ... if the object were thrown upwards with an upwards acceleration, that would not be the case would it?

Chestermiller
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Nov6-12, 07:36 AM
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Basic Kinematics Question

When an object is thrown upwards, the acceleration is still 9.8 m/s2 downwards, but there is now an initial upward velocity imposed on the object which was simply not present in the case where the object was initially at rest and allowed to drop.
Simon Bridge
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Nov7-12, 12:46 AM
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Quote Quote by Chestermiller View Post
When an object is thrown upwards, the acceleration is still 9.8 m/s2 downwards, but there is now an initial upward velocity imposed on the object which was simply not present in the case where the object was initially at rest and allowed to drop.
Not the situation described though - here there was an initial speed.


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