# Pit dropping time difference

1. Sep 25, 2011

### physicsgurl12

1. The problem statement, all variables and given/known data
An object is dropped from rest into a pit and accelerates due to gravity at roughly 10m/s^2. It hits the ground in 5 seconds. A rock is then dropped from rest into a second pit and hits the ground in ten seconds.How much deeper is the second pit. no air resistance.

2. Relevant equations

not used

3. The attempt at a solution
im pretty sure the answer is 4 times.
but im not 100%. is it sorta like braking from one velocity to stop then a second(half as much) to stop and finding the distance it takes to stop.because thats four times as long.

2. Sep 25, 2011

### I like Serena

Hey pg!

Let's get back to formulas.
Do you know a relevant formula from which you could calculate the depth of the pit?

3. Sep 25, 2011

### physicsgurl12

what if i do a=d/t/t
so it would be 10m/s^2=d/5s/5s
250m=d

same formula
10m/s^2=d/10s/10s
1000m=d

4. Sep 25, 2011

### I like Serena

Hmm, "a=d/t/t" is not a correct formula.

You either should have $a = {d^2x \over dt^2}$,
or what would suit your problem:
$$d = {1 \over 2}g t^2$$

5. Sep 25, 2011

### physicsgurl12

but a=d/t/t is eaual to a=v/t

6. Sep 25, 2011

### I like Serena

Not quite.
v=d/t gives you the "average" speed during the entire drop.
The actual speed starts at zero and increases to some maximum.
Assuming the acceleration is constant, the corresponding acceleration is actually a=2 d/t/t.

In other words, you cannot just use "a=v/t" or "v=d/t".
You should use: "d=d0 + v0 t + (1/2)a t^2" (assuming acceleration a is constant).
And: "v=v0 + a t" (again assuming a is constant).

7. Sep 25, 2011

### physicsgurl12

okay well i used this and still got four.

8. Sep 25, 2011

### I like Serena

Yep!

(Sorry to drag this out, but now you did it with the proper formula. )

9. Sep 25, 2011

### physicsgurl12

haha its okay as long as i understand.i wanna do well on my test.:)