Initial Velocity of a falling object

[eNathan]

Hello everyone

I know that $$\sqrt { \frac {2d} {g} }$$ but what about including initial velocity?

Thanks in advanced

"Im the master of time" -- Eiffel 65

 

[robphy]

I know that $$\sqrt { \frac {2d} {g} }$$ but what about including initial velocity?

that _what_ is $$\sqrt { \frac {2d} {g} }$$?
It's $$t= \sqrt { \frac {2d} {g} }$$.
Where did that equation come from?

 

[Doc Al]

$$y = y_0 + v_0 t - \frac {g t^2}{2}$$

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; $y_0$ is the initial height; $v_0$ is the initial speed.

 

[eNathan]

robphy, we all know what I meant.

Thanks Doc_Al!

 

[RENATO]

$$y = y_0 + v_0 t - \frac {g t^2}{2}$$

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; $y_0$ is the initial height; $v_0$ 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?

 

[Doc Al]

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?

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]

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 +volt +yo, yo being the initial height. What is not clear to me is why is t (time) the same throughout the equation.

 

[Doc Al]

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

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]

Thank you so much. I do understand now.

 

[cstoos]

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.

 

[Doc Al]

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 have it backwards. That equation only deals with vertical motion (in the y direction). The v₀ in that equation is just the y-component of the initial velocity.

 

[RENATO]

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 +volt +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.

 

[Doc Al]

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

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.)

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.

 

[RENATO]

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.

 

[RENATO]

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

 

[RENATO]

or is it 3 seconds, now I am guessing.

 

[RENATO]

Another way of solving, which I am not sure is:

d =volt + (0.5)at^2
50= (-15)t + 16.1 t^2
t= 2.28 seconds

 

[Doc Al]

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

Looks fine to me.

or is it 3 seconds, now I am guessing.

What's the point of guessing?

Another way of solving, which I am not sure is:

d =volt + (0.5)at^2
50= (-15)t + 16.1 t^2
t= 2.28 seconds

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?

 

[RENATO]

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!

 

[RENATO]

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?

 

[Doc Al]

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

Yes.

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

Explain your difficulty. Hint: Just because the force is the same doesn't mean the effect of the force is the same.

 

[RENATO]

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.

 

[Doc Al]

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?

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 are internal to the system). That means the total momentum of both vehicles together doesn't change. Is that your question?

 

[RENATO]

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.

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

 

[Doc Al]

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.

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]

I am really thick headed.

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

 

[Doc Al]

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

Sure. And the volkswagon will hit the truck with the same force in the opposite direction.

 

[RENATO]

I can understand when a vehicle hit a wall. The vehicle will have an equal and opposite reaction equal to the force of the car.

 

[RENATO]

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?

 

[Doc Al]

I can understand when a vehicle hit a wall. The vehicle will have an equal and opposite reaction equal to the force of the car.

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

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