# Homework Help: Calculate when displacement are equal

1. Feb 28, 2013

### ahmedb

1. The problem statement, all variables and given/known data

One stone is dropped from the the top of a tall cliff, and a second stone with the same mass is thrown vertically from the same cliff with a velocity of 10.0 m/s [down], 0.50seconds after the first. Calculate the distance below the top of the cliff at which the second stone overtakes the first?

2. Relevant equations
v=0
d= -1/2(a)(t)^2

v=10
d=10(t)-1/2(a)(t)^2

3. The attempt at a solution

we have two unknowns, how can we solve it?!?!

2. Feb 28, 2013

### SammyS

Staff Emeritus
The time in this equation should be (t - 0.50), if t is the time in the previous equation .
Two equations in two unknowns.

3. Feb 28, 2013

### ahmedb

i still don't get the answer, can you explain the steps please

4. Feb 28, 2013

### SammyS

Staff Emeritus
First rewrite the second equation,

d=10(t)-1/2(a)(t)2 ,

using (t - 0.5) in place of t. (This is because the second stone is in the air for 1/2 second less time than the first stone.)

Then "FOIL" the (t - 0.5)2, multiply out everything in the expression and then collect terms.

See what you have then.

5. Feb 28, 2013

### ahmedb

so after you foil, you make the displacement equal to each other, and then solve for t, and then plug "T" into the original equation?

6. Feb 28, 2013

### SammyS

Staff Emeritus
Yes.

7. Feb 28, 2013

### ahmedb

can you also do: d= -1/2(a)(t+.5)^2

cas the ffirst stone is .5 seconds in air longer than the second stone

8. Feb 28, 2013

### SammyS

Staff Emeritus
You can do either

t for the first stone and (t - 0.5) for the second stone,

OR

(t - 0.5) for the first stone and t for the second stone,

but not both (t + 0.5) and (t - 0.5).

9. Feb 28, 2013

### ahmedb

don't you mean "t for the first stone and (t "+" 0.5) for the second stone,"?

10. Feb 28, 2013

### SammyS

Staff Emeritus
No. No matter how you express the time for each stone the time for the second stone is always 0.5 s less than the time for the first stone.

Therefore, the time for the first stone is always 0.5 s more than the time for the second stone.

11. Feb 28, 2013

### ahmedb

(t - 0.5) for the first stone and t for the second stone,

for that cas the time for the first one is 5 less than the time for the second stone.

so is this correct "(t + 0.5) for the first stone and t for the second stone,"?

12. Feb 28, 2013

### SammyS

Staff Emeritus
Then, while the first stone falls for say 1 second, the the second stone would fall for 1 + 0.5 seconds. That's just not correct.

13. Feb 28, 2013

### ahmedb

ohh lol, now i get it, thanks :D