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Initial Velocity of a falling object

by eNathan
Tags: falling, initial, object, velocity
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eNathan
#1
May9-05, 06:45 PM
P: 352
Hello everyone

I know that [tex] \sqrt { \frac {2d} {g} } [/tex] but what about including initial velocity?

Thanks in advanced

"Im the master of time" -- Eiffel 65
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robphy
#2
May9-05, 07:01 PM
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Quote Quote by eNathan
I know that [tex] \sqrt { \frac {2d} {g} } [/tex] but what about including initial velocity?
that _what_ is [tex] \sqrt { \frac {2d} {g} } [/tex]?
It's [tex]t= \sqrt { \frac {2d} {g} } [/tex].
Where did that equation come from?
Doc Al
#3
May9-05, 07:03 PM
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[tex]y = y_0 + v_0 t - \frac {g t^2}{2}[/tex]

This is the kinematic equation describing a falling body (ignoring air resistance). Here "up" is positive and "down" is negative; "y" is the position after t seconds; [itex]y_0[/itex] is the initial height; [itex]v_0[/itex] is the initial speed.

eNathan
#4
May10-05, 09:20 AM
P: 352
Initial Velocity of a falling object

robphy, we all know what I meant.

Thanks Doc_Al!
RENATO
#5
May19-10, 07:18 AM
P: 18
Quote Quote by Doc Al View Post
[tex]y = y_0 + v_0 t - \frac {g t^2}{2}[/tex]

This is the kinematic equation describing a falling body (ignoring air resistance). Here "up" is positive and "down" is negative; "y" is the position after t seconds; [itex]y_0[/itex] is the initial height; [itex]v_0[/itex] is the initial speed.
I am still not clear why the time of the initial velocity (going up) is equal to the t in 1/2gt^2. Will someone enlighten me?
Doc Al
#6
May19-10, 07:30 AM
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Quote Quote by RENATO View Post
I am still not clear why the time of the initial velocity (going up) is equal to the t in 1/2gt^2. Will someone enlighten me?
I'm unclear what you are asking. What do you mean by "time of the initial velocity"? Please rephrase your question. What problem are you trying to solve?
RENATO
#7
May19-10, 07:43 AM
P: 18
If a person is standing on top of the cliff, throws the ball upwards at 15 ft/sec from an initial height of 50 ft. How high is the rock after 2 seconds. Also, what is the total time when it hits the ground at 50 ft below.

I know I can solve the problem by following the formula y=1/2 at^2 +Vot +yo, yo being the initial height. What is not clear to me is why is t (time) the same throughout the equation.
Doc Al
#8
May19-10, 07:54 AM
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Quote Quote by RENATO View Post
What is not clear to me is why is t (time) the same throughout the equation.
The time (t) in that equation is a parameter that continually changes. t = 0 is the moment when the ball is first thrown. That equation tells you how the position of the ball changes as a function of time, where time is measured from the moment the ball was thrown.
RENATO
#9
May19-10, 08:00 AM
P: 18
Thank you so much. I do understand now.
cstoos
#10
May19-10, 08:09 AM
P: 64
That equation still will not give you the correct answer. That is for if the initial velocity is orthaginol to gravity. You are not accounting for the distance traveled in the positive y direction.

You need velocity to start out positive at a decreasing rate, and end up negative at an increasing rate. Either work it piecewise find a path function.
Doc Al
#11
May19-10, 08:16 AM
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Quote Quote by cstoos View Post
That equation still will not give you the correct answer. That is for if the initial velocity is orthaginol to gravity. You are not accounting for the distance traveled in the positive y direction.
You have it backwards. That equation only deals with vertical motion (in the y direction). The v0 in that equation is just the y-component of the initial velocity.
RENATO
#12
Jun17-10, 02:08 PM
P: 18
[QUOTE=RENATO;2723823]If a person is standing on top of the cliff, throws the ball upwards at 15 ft/sec from an initial height of 50 ft. How high is the rock after 2 seconds. Also, what is the total time when it hits the ground at 50 ft below.

I know the formula y=1/2 at^2 +Vot +yo, yo being the initial height. What is not clear to me is why is t (time) the same throughout the equation.

Will someone solve for the height after 2 seconds and the total time so I would know if my answers are correct.
Doc Al
#13
Jun17-10, 02:18 PM
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Quote Quote by RENATO View Post
Will someone solve for the height after 2 seconds and the total time so I would know if my answers are correct.
Why don't you show what you did and we'll take a look at your work? (As I thought was explained before, time is just a parameter. That formula gives the position as a function of time.)

In one case, the time is given and you'll calculate the position. In the other case you know the final position and you have to solve for the time. You use the same equation for both parts.
RENATO
#14
Jun17-10, 02:28 PM
P: 18
I can only solve for the height after 2 seconds which is 15.6 ft, but I do not know how to solve for the total time.
RENATO
#15
Jun17-10, 02:42 PM
P: 18
Is the following correct:
-50 = Vot -32.2/2(t^2)
-50 = 15t -16.1t^2
t = 2.28
RENATO
#16
Jun17-10, 02:50 PM
P: 18
or is it 3 seconds, now I am guessing.
RENATO
#17
Jun17-10, 03:11 PM
P: 18
Another way of solving, which I am not sure is:

d =Vot + (0.5)at^2
50= (-15)t + 16.1 t^2
t= 2.28 seconds
Doc Al
#18
Jun17-10, 03:43 PM
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Quote Quote by RENATO View Post
Is the following correct:
-50 = Vot -32.2/2(t^2)
-50 = 15t -16.1t^2
t = 2.28
Looks fine to me.

Quote Quote by RENATO View Post
or is it 3 seconds, now I am guessing.
What's the point of guessing?

Quote Quote by RENATO View Post
Another way of solving, which I am not sure is:

d =Vot + (0.5)at^2
50= (-15)t + 16.1 t^2
t= 2.28 seconds
This is equivalent to the first method--you just multiplied both sides by -1. (It's the same equation.)

So it seems that you understand how to solve for the time after all. Or do you still have a question?


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