Physics: How to solve this problem and others similar to it?

Question

I am confused on how to solve this problem and problems similar to it. When I try to do them I always panic and forget how to solve them. Can anyone help me?

A rock is dropped down a canyon and hits 15 sec later?
-how deep is it?
-how fast is it when it hits?

The equations that we are supposed to use are:
V=X/T
Vf=Vi+AT
Vf^2=Vi^2+2AX
X=ViT+1/2AT^2
V=velocity, X=distance, T=time, Vf= final velocity, Vi= initial velocity, A=acceleration

Quaternion · Accepted Answer

This is going wayyyyyy back, but here's what I know... 

you know T (15s), Vi (0 because you drop the rock, not throw it), A (9.8 m/sec2 if memory serves).  Now you just need to use the two equations to solve for Vf and X.  Luckily, since Vi is zero, that gets rid of the ViT and Vi terms, so Vf = AT and X = 1/2 AT^2.  That should be easy enough...

? · Answer

Tips.....

V=X/T
Only use this if the velocity is uniform, meaning the velocity doesn't change. If acceleration is involved, then velocity changes and you can rule this one out.

If it's falling, then you know it's acceleration is A = 9.8, caused by gravity.
For falling objects, unless they give you an initial velocity, assume it starts from rest, Vi=0

For extra practice, you can also use this equation to find it's depth
Vf^2=Vi^2+2AX –– solve for X
(Vf^2 - Vi^2) / 2A = X
(147^2 - 0) / 2(9.8) = X
1102.5 = X

? · Answer

Use the second equation to find the Final Velocity

Vf=0+9.8(15)
Vf=147m/s

And the last equation can tell you depth
X=(0)15+1/2(9.8)15^2
X=1102.5 m

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Physics: How to solve this problem and others similar to it?

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