BR Physics Ch 1 Passage 3 Question 20

[jw1985]

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Hi,

I am confused with the following question because we are supposed to use the following Kinematic equation to arrive at the answer.

Vt^2 = Vo^2 + 2a(delta Y)

I am confused because I don't understand how the height is proportional to the initial speed. Looking at the equation, when Isolating Vo, isn't delta Y inversely proportional?

I have attached the question. The answer is C.

Thanks in advance for all your help!

 

[Gauss44]

Hi,

I am confused with the following question because we are supposed to use the following Kinematic equation to arrive at the answer.

Vt^2 = Vo^2 + 2a(delta Y)

I am confused because I don't understand how the height is proportional to the initial speed. Looking at the equation, when Isolating Vo, isn't delta Y inversely proportional?

I have attached the question. The answer is C.

Thanks in advance for all your help!

It stands to reason that the greater the initial velocity, the greater the height will be. You have to weigh that against the negative sign that I think is causing confusion.

If I were to try to rationalize the negative sign, it is probably indicative of direction.

 

[Romz]

son of a bitch lol


I got it wrong too. Read the answer C.

As we increase the INITIAL velocity we throw an object, it's height will be Greater!

That's why we see the curve for maximum height shoot up as we see vo increase.

This leaves a and c. V is on the second degree curve compared to height y, explaining it's parabolic like trajectory.

 

[jw1985]

It stands to reason that the greater the initial velocity, the greater the height will be. You have to weigh that against the negative sign that I think is causing confusion.

If I were to try to rationalize the negative sign, it is probably indicative of direction.

Ah, I see. Then I guess I would arrive with the correct signage if I gave acceleration a negative value.

Makes sense.

and @Domenic, thanks for pointing out that it's a secondary degree type curve.