Determine velocity of falling object after certain time

1. Sep 12, 2004

soccerjayl

Need a little help on Kinematics.

Say for instance, an object was dropped by a steadily rising body. To determine the velocity of the falling object after a certain time would be V(f)=(-9.81)(t). But if one were to find the distance between the falling object and the rising body, how could you do it?

I tried by finding the change in displacement (delta x=(1/2)(-9.81)(t)^2) and added (t)(m/s rate of rising body). I've been told this is incorrect. If incorrect, how can i approach any problems like that in the future?

2. Sep 12, 2004

Tide

You need to account for the fact that the object is initially rising when it is released.

3. Sep 12, 2004

soccerjayl

meaning....

i thought adding the time*rate would be enough

e.g. displacement -35 at 7 seconds (rising body is rising at .5 m/sec)

distance between falling body and rising at 7 seconds

-35 + [(7)(.5)]

what exactly do u mean? what other variable will help me figure out examples like this?

4. Sep 12, 2004

$$v = v_0 - gt[/itex] 5. Sep 12, 2004 munky99999 i didnt quite catch all the info from the posts. Looks like your trying to do it all in one equation. 6. Sep 12, 2004 Cyrus using the equation x = xint + v int *t + 1/2 a t^2 , you would obtain the distance the object has fallen realive to a point in space that is not moving. However, you have to remember that as time goes by, the steadily rising body will be moving up. So you have to find a way to factor this in. perhaps if you just add this upwards velocity* time to the equation, you will get the total distance from the body to the projectile. I havent thought this through too much so it may be wrong. It seems to me what you explained is right. The initial velocity can be considered the upward velocity of the body. Id like to know what the correct anwser is though. I would think the critical thing to be to include the intial height of the body the moment its released as the xinitial term. Err on second thought cancel that idea. I think I see what is wrong with using that equation. If you use that equation, it will tell you how high you are above the ground at any instant in time. It wont tell you how high you are realtive to the object you feel from. That being the case you should be able to solve it very easily now. You can calculate your height above the ground using your equation. But now you have to also calculate how high the body went up in that amount of time. Once you know how high up it went, you can subtract its distance from the ground at that point, and you should be able to easily calculate the distance between the two. Last edited: Sep 12, 2004 7. Sep 12, 2004 Cyrus Here is the position of the projectile as a fucntion of time (IMPORTANT NOTE: it is the position of the projectile RELAIVE TO THE GROUND NOT THE BODY MOVING UP) [tex] x = x_0 + v_0 *t + \frac{a t^2} {2}$$

( I Think the initial velocity will not be zero in this case, as it is moving upwards while its being dropped, so there might be an initial velocity to it, I think).

And the bodys position will be the sum of its initial height and the product of the velocity times the distance it has traveled in that amount of time:

$$x = v_b_o_d_y * t + x_0$$

Notice $$x_0$$ is the same initial height in both equations since both start off from the same height.

So to find the difference in heigh, you have to subtract the two distances, or:

$$difference-in-heights = (v_b_o_d_y * t + x_0) - (x_0 + v_0 *t + \frac {a t^2} {2} )$$

It seems like all this junk cancels out leaving you with:

$$1/2 * a * t^2$$

Last edited: Sep 13, 2004
8. Sep 13, 2004

soccerjayl

im gonna come out, im still having trouble...

mainly its this problem...do not do the problem, but just see if im doin anything wrong

A physics student throws a softball straight up into the air. The ball was in the air for a total of 5.96 s before it was caught at its original position. The initial velocity is 29.2 m/sec

How high did it rise?
I found it to be 131 m

0=(29.2)-(-9.81)t
t=2.98

x=(29.2)(2.98) - (1/2)(-9.81)(2.98)^2
x=131 m

What did i do wrong?

9. Sep 13, 2004

soccerjayl

arg, thats the problem unrelated to the subject. If anyone can help in it, plese do so, but here is the correct problem related to what ive talked about.

A small fish is dropped by a pelican that is rising steadily at 0.40 m/s. (Assume the positive direction is upward.) After 3.5 seconds, velocity is -34.3 m/s

How far below the pelican is the fish after 3.5 s?

I got it to be -34.7 by how its been described...

can someone just follow the equations and not necessarily provide the answer

10. Sep 13, 2004

Cyrus

V = v0 + at you have two minus signs. its + at. a is -9.8 it should be -9.8 t not - - 9.8t

11. Sep 13, 2004

soccerjayl

yes!!

disregard last messageg

i found the answer: 61.1 m

If anyone could help me on my post at 1:03 AM EAST, that would be great

but thanks again for what u did for me in the last problem

12. Sep 13, 2004

Cyrus

was it the minus sign? I hate when that happens to me.

13. Sep 13, 2004

soccerjayl

yes it was

thank u guys so much tonight

u saved my night

thanks again

14. Sep 13, 2004

Cyrus

how about the other problems you get them solved yet?

15. Sep 13, 2004

Cyrus

Im very interested to know the problem and anwser abotu the one concerning the body that is moving upwards as the thing drops down

16. Sep 13, 2004

Tide

Jay's original conjecture was (almost) correct! I misinterpreted his original query.

Just for fun how about doing it from the perspective of the rising body! The distance the object falls is $s_{object} = g t^2 /2$ since the body's reference frame is not accelerating. As a bonus, you could find out how long it takes to reach the ground by noting the distance to the ground is $s_{ground} = h + v_{body}t$ where h is the initial distance to the ground. Set these distances equal and solve for t.

17. Sep 13, 2004

Cyrus

Cant he just use my formula which basically is the same as yours tide? im just curious if i derived it correctly or not.

Last edited: Sep 13, 2004
18. Sep 13, 2004

Tide

Cyrus,

Yours looks good to me - nice work.

19. Sep 13, 2004

soccerjayl

here you go:

A small fish is dropped by a pelican that is rising steadily at 0.40 m/s. (Assume the positive direction is upward.)
(a) After 3.5 s, what is the velocity of the fish?
(b) How far below the pelican is the fish after 3.5 s?