Solving Kinematics Problem: 2 Stones Dropped from 60-m Cliff

[superstorebug]

I'm fairly new with this so I hope I've posted at the right place!


I don't understand how to go about doing the following kinematics problem:

Two stones are dropped from the edge of a 60-m cliff, the second stone 1.6s after the first. How far below the top of the cliff is the second stone when the separation between the two stones is 36 m?

I hope you can help because I have a quiz on this stuff tomorrow! Thanks in advance.

ANy help will be greatly appreciated.

 

[superstorebug]

The answer is 10.9 meters...
but I'm still confused as how to get there...
I've used:
-g as my acceleration
1.6 as my change in time
but I'm not sure what to use for intial and final positon

...is it possible to do this using just one equation?

 

[chemhelper]

Hi,

Do you know of any of the kinematic equations that may be relevant here (maybe one that deals with displacement and there are only a few)?

And no, you cannot use a single equation ( you are going to have to use two equations, though they are both the same type of equation kinematic equation as noted from above)

 

[superstorebug]

I'm trying to use the following kinematics equation:

x(t)=x +vt +(1/2)at^2

 

[chemhelper]

Thats correct. Your flaw comes when you try to use 1.6 as the change in time as time is a variable.

Ok, I lied when I said that there you use only one form of an equation, there are two.

The one you defined is the one you should use first. If you set up two equations, you'll end up with X1 and X2.

However, the second stone is exactly 36 m behind the first. So you will get X2=X1 - 36.

Also, remember that the second stone was thrown 1.6 seconds after the first (t2=t1-1.6)