# Object Falling off of Table

1. Oct 28, 2009

### merceb49

A toy car runs off the edge of a table that is 1.325 m high. The car lands 0.425 m from the base of the table.

How long did it take to hit the ground?
How fast was the car going?

My tutor even had trouble with this, without a speed this is no solvable, at least to me. Suggestions on formulas? And you cant answer the second one without the first one, so any suggested formulas or words of wisdom?

2. Oct 28, 2009

### leright

Calculate the time it takes the car to hit the ground first. Use the formula 0.5at^2 = distance. You know the distance the car fell is 1.325 m.

The speed of the car is simply (0.425m)/time.

3. Oct 28, 2009

### merceb49

So i do not have the time, or the acceleration. i cant calculate the acceleration because i dont have the mass of the car. I understand how to get the speed but I dont have the time so I dont know how to solve it. How can I find the acceleration?

4. Oct 28, 2009

### XanziBar

I just dropped my pen, what was its acceleration while falling?

5. Oct 28, 2009

### pr0blumz

What's the acceleration due to gravity?

6. Oct 28, 2009

### merceb49

wow i'm stupid. but i tried this all
.5(9.8)t^2=1.325 and I got around 0.5. And this is still incorrect, should I not use the distance it fell horizontally or should I add both of the distances together?

7. Oct 28, 2009

### pr0blumz

t=sqrt 2h/a

8. Oct 28, 2009

### merceb49

with that formula i got 0.3, assuming i did it correctly (t = sqrt of 2(.425)/9.8)
Is that not a ridiculous number?

9. Oct 28, 2009

### pr0blumz

h=1.325, not .425

10. Oct 28, 2009

### pr0blumz

Also, v_f^2 = v_i^2 +2ax

11. Oct 28, 2009

### merceb49

sqrt of 2(1.325)/9.8=t=.5=fast?

12. Oct 28, 2009

### pr0blumz

t = time, not velocity

13. Oct 28, 2009

### merceb49

yeah im trying to answer the first question, and i thank you for coping with my confusion... i understand the second question, i have to find how long it took to find the velocity but i still dont think .5 is the correct time

14. Oct 28, 2009

### pr0blumz

a) v (vertical speed) = 0
vt+(1/2)*gt^2=s
t = sqrt (2s/g)

b)b) v = horizontal speed
vt+(1/2)at^2 = s
a = 0
v = s/t

15. Oct 29, 2009

### Bloodthunder

Does the answer key say otherwise? If not 0.520s is the correct time. Don't believe it, drop something from 1.3 m up and see how long it takes to fall to the ground. The initial velocity is only due to the distance the car moves in the horizontal position as there is no initial vertical velocity. The initial velocity, which is the horizontal velocity, is 0.818 m/s.

And get a better tutor.

16. Oct 29, 2009

### pr0blumz

I don't have my calc. so i can't verify your answers, sorry.

17. Oct 29, 2009

### pr0blumz

Yes, you need a better tutor. I don't mind helping though.

18. Oct 29, 2009

### Bloodthunder

It should be right, I think, I did my calculations similar to what you posted earlier.

19. Oct 29, 2009

### merceb49

i was keying it incorrectly, bloodthunder your correct. thank you for all the help on this problem

20. Oct 29, 2009

### pr0blumz

i guess your frustration got the best of ya.