What is the final V in the vertical direction?

AI Thread Summary
The discussion revolves around solving a physics problem related to vertical velocity. The user initially calculates the final vertical velocity (Vy) using two equations, arriving at conflicting results. The first equation yields 32.83 m/s, while the second gives zero due to misinterpretation of the variables involved. A key correction is noted that the first equation should use Δy instead of time (t), clarifying the calculation of Vy. The user expresses a broader struggle with understanding physics concepts, despite proficiency in other subjects.
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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
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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