What is the final V in the vertical direction?

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Homework Help Overview

The discussion revolves around a physics problem involving kinematics, specifically the calculation of final vertical velocity (Vy) for an object thrown horizontally from a height. Participants are exploring the application of kinematic equations to determine the correct value of Vy and are grappling with the implications of initial conditions and sign conventions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using the equations Vy2 = Vyo2 + 2ayt and y = 1/2(Vyo + Vy)t to find Vy. There is confusion regarding the initial vertical velocity (Vyo) and how to treat the displacement (Δy) in calculations. Some participants question the interpretation of results when considering the signs of values involved.

Discussion Status

The discussion is ongoing, with participants providing insights and corrections regarding the equations used. There is an acknowledgment of errors in the application of the equations, particularly in the context of the displacement and initial conditions. Multiple interpretations of the problem are being explored, and some guidance has been offered regarding the correct use of the equations.

Contextual Notes

Participants express a broader struggle with understanding physics concepts compared to other subjects, indicating a potential underlying issue with conceptualizing physical problems. There is mention of specific numerical values and the need for clarity on the assumptions made regarding the motion of the object.

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Homework Statement


upload_2019-2-13_21-47-2.png


Homework Equations


Vy2 = Vyo2 + 2ayt
y = 1/2(Vyo + Vy)t

The Attempt at a Solution


So I have "solved" this problem with both the equations above.

Using the Vy2 equation I get the result of 32.83m/s.
Using the other equation I get a result of zero, because Δy = (0 - 55) = -55. However, if I consider that 55m a positive value, then I get 32.83m/s... same as first equation.

I knew Vyo = 0m/s because the initial velocity would only be in the x-direction since the ball "is thrown horizontally". To prove it I did the equation Δy = Vyot + 1/2ayt and got a result of 0m/s.

So, where am I going wrong?

I actually have a much bigger problem than just this one I'm posting. For some reason I cannot wrap my head around physics problems. I can do calculus, logic classes, philosophy, etc, but for some reason cannot make my mind look at physic's problems in the right way. Conceptually, all of these kinematic questions seem easy, but I am spending WAY more time on these problems than should be necessary. If you have an answer for this conundrum I welcome your responses...
 

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I don't understand how you get ##V_y=0## using the second equation.

Isnt it ##-55=\frac{1}{2}(0+V_y)3.35\Rightarrow V_y=\frac{-2\cdot55}{3.35}=-32.83##?
 
Lapse said:

Homework Statement


[ ATTACH=full]238714[/ATTACH]

Homework Equations


Vy2 = Vyo2 + 2ayt
y = 1/2(Vyo + Vy)t

The Attempt at a Solution


So I have "solved" this problem with both the equations above.

Using the Vy2 equation I get the result of 32.83m/s.
Using the other equation I get a result of zero, because Δy = (0 - 55) = -55. However, if I consider that 55m a positive value, then I get 32.83m/s... same as first equation.

I knew Vyo = 0m/s because the initial velocity would only be in the x-direction since the ball "is thrown horizontally". To prove it I did the equation Δy = Vyot + 1/2ayt and got a result of 0m/s.

So, where am I going wrong?

I actually have a much bigger problem than just this one I'm posting. For some reason I cannot wrap my head around physics problems. I can do calculus, logic classes, philosophy, etc, but for some reason cannot make my mind look at physic's problems in the right way. Conceptually, all of these kinematic questions seem easy, but I am spending WAY more time on these problems than should be necessary. If you have an answer for this conundrum I welcome your responses...
You have an error in the first equation you list
Vy2 = Vyo2 + 2ayt​

You should have Δy rather than t , so it should be:
Vy2 = Vyo2 + 2ay(Δy) ,​
which is probably what you actually used if you got 32.83m/s for Vy.

Notice that when you solve an equation such as ##v_y^2 = 2(-9.8)(-55)## by taking the square root of both sides, the answer is:
##v_y = \pm \sqrt{1078} \approx \pm 32.83 ~. ##​

You then need to choose the sign consistent with the situation
 

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