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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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- Thread starter eNathan
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In summary, both the Volkswagen and the large truck experience the same force of impact in a head-on collision. This is based on Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. Therefore, the magnitude of force experienced by both vehicles is equal.

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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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eNathan said: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?

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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.

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robphy, we all know what I meant.

Thanks Doc_Al!

Thanks Doc_Al!

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Doc Al said:

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?

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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 said: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?

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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.

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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 said:What is not clear to me is why is t (time) the same throughout the equation.

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Thank you so much. I do understand now.

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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.

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You have it backwards. That equationcstoos said: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.

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RENATO said: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.

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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.)RENATO said:Will someone solve for the height after 2 seconds and the total time so I would know if my answers are correct.

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

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Is the following correct:

-50 = Vot -32.2/2(t^2)

-50 = 15t -16.1t^2

t = 2.28

-50 = Vot -32.2/2(t^2)

-50 = 15t -16.1t^2

t = 2.28

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or is it 3 seconds, now I am guessing.

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d =Vot + (0.5)at^2

50= (-15)t + 16.1 t^2

t= 2.28 seconds

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Looks fine to me.RENATO said:Is the following correct:

-50 = Vot -32.2/2(t^2)

-50 = 15t -16.1t^2

t = 2.28

What's the point of guessing?RENATO said:or is it 3 seconds, now I am guessing.

This is equivalent to the first method--you just multiplied both sides by -1. (It's the same equation.)RENATO said:

d =Vot + (0.5)at^2

50= (-15)t + 16.1 t^2

t= 2.28 seconds

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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doc al said:looks fine to me.

What's the point of guessing?

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?

thanks, that's it. It is nice to have a person like you always ready to help!

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Momentum Problem:

A large truck and a Volkswagen have a head-on collision.

Which vehicle experiences the greatest force of impact?

Answer: Both the Volkswagon and the large truck encounter the same force.

Does it mean that the magnitude of force is the same for both vehicles?

I am having difficulty understanding this even if I apply Newton’s third law

of motion.

Can someone help me on this?

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Yes.RENATO said:Does it mean that the magnitude of force is the same for both vehicles?

Explain your difficulty. Hint: Just because the force is the same doesn't mean theI am having difficulty understanding this even if I apply Newton’s third law

of motion.

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My difficulty is this:

I know that the Volkswagen will move with more acceleration after the impact because of it's smaller mass.

But, I am always thinking of the second law of motion (F=ma), how do we determine the resultant force for the two vehicles?

Doc Al - thanks for being there for confused persons like me.

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When the truck and car collide the net force on the total system of both vehicles is zero--since the forces they exert on each other are equal and opposite (they areRENATO said:My difficulty is this:

I know that the Volkswagen will move with more acceleration after the impact because of it's smaller mass.

But, I am always thinking of the second law of motion (F=ma), how do we determine the resultant force for the two vehicles?

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Suppose the truck has a force of 1000 lbs and the Volkswagen has 100 lbs. Please tell me the net force of the system.

Also, will the truck have an opposite reaction of 1000 lbs? And the Volkswagen 100lbs?

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I thought you were talking about the impact force due to their collision? In which case, they must exert equal and opposite forces on each other. That's Newton's 3rd law. You can't have one exert a different force than the other.RENATO said:Please bear with me.

Suppose the truck has a force of 1000 lbs and the Volkswagen has 100 lbs. Please tell me the net force of the system.

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I am talking about the force. Won't the truck hit the volkswagen and push the latter?

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Sure. And the volkswagon will hit the truck with the same force in the opposite direction.RENATO said:I am talking about the force. Won't the truck hit the volkswagen and push the latter?

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Doc Al said:Sure. And the volkswagon will hit the truck with the same force in the opposite direction.

What about the truck? Will it hit the Volkswagen with equal opposite reaction?

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When the car hits the wall they exert equal and opposite forces on each other.RENATO said:

That doesn't just apply to walls. Whenever

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Did you get my last question?

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Doc Al said:When the car hits the wall they exert equal and opposite forces on each other.

That doesn't just apply to walls. Wheneveranytwo things hit each other, they exert equal and opposite forces on each other.

Doc Al - thanks for your patience. I do understand now and I can sleep soundly.

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Of course. If the truck hits the volkswagon, then the volkswagon hits the truck. They exert equal and opposite forces on each other, just likeRENATO said:What about the truck? Will it hit the Volkswagen with equal opposite reaction?

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Doc Al said:Of course. If the truck hits the volkswagon, then the volkswagon hits the truck. They exert equal and opposite forces on each other, just likeanytwo things that hit each other.

Thank you so much!

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