Initial Velocity of a falling object

1. May 9, 2005

eNathan

Hello everyone

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

"Im the master of time" -- Eiffel 65

2. May 9, 2005

robphy

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

3. May 9, 2005

Staff: Mentor

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

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.

4. May 10, 2005

eNathan

robphy, we all know what I meant.

Thanks Doc_Al!

5. May 19, 2010

RENATO

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?

6. May 19, 2010

Staff: Mentor

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?

7. May 19, 2010

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

8. May 19, 2010

Staff: Mentor

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.

9. May 19, 2010

RENATO

Thank you so much. I do understand now.

10. May 19, 2010

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.

11. May 19, 2010

Staff: Mentor

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.

12. Jun 17, 2010

RENATO

13. Jun 17, 2010

Staff: Mentor

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.

14. Jun 17, 2010

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.

15. Jun 17, 2010

RENATO

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

16. Jun 17, 2010

RENATO

or is it 3 seconds, now I am guessing.

17. Jun 17, 2010

RENATO

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

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

18. Jun 17, 2010

Staff: Mentor

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?

19. Jun 17, 2010

RENATO

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

20. Jul 1, 2010

RENATO

Re: MOMENTUM

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?