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Initial Velocity of a falling object 
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#1
May905, 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 


#2
May905, 07:01 PM

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It's [tex]t= \sqrt { \frac {2d} {g} } [/tex]. Where did that equation come from? 


#3
May905, 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. 


#4
May1005, 09:20 AM

P: 352

Initial Velocity of a falling object
robphy, we all know what I meant.
Thanks Doc_Al! 


#5
May1910, 07:18 AM

P: 18




#6
May1910, 07:30 AM

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#7
May1910, 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. 


#8
May1910, 07:54 AM

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#9
May1910, 08:00 AM

P: 18

Thank you so much. I do understand now.



#10
May1910, 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. 


#11
May1910, 08:16 AM

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#12
Jun1710, 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. 


#13
Jun1710, 02:18 PM

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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. 


#14
Jun1710, 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.



#15
Jun1710, 02:42 PM

P: 18

Is the following correct:
50 = Vot 32.2/2(t^2) 50 = 15t 16.1t^2 t = 2.28 


#16
Jun1710, 02:50 PM

P: 18

or is it 3 seconds, now I am guessing.



#17
Jun1710, 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 


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