## catsrule332 3 years ago A rock is pushed off of a cliff. If air resistance can be neglected, how far will the rock have fallen after 4s? Explain.

1. loophole144

acceleration due to gravity=9.8 m/s^2

2. Callum29

If the rock is pushed off (supposing the push is in the horizontal direction so it does not affect the motion) then the initial vertical velocity is 0ms^-1. gravity is 9.8ms^-2 and we have that: $S= ut + at^2 = 0 + 9.8(4)^2 = 156.8m$

3. catsrule332

I understand that but I'm not sure I set the equation up right. I used: $DeltaX=1/2at ^{2}$

4. Callum29

sorry, $S= ut + \frac{1}{2}at^2 = 4.9\times 16 = 78.4m$

5. catsrule332

Okay so the 1/2 is right?

6. Callum29

that is the formula one should use...

7. Callum29

Yeah, the 1/2 is correct there in that formula. $S= ut + \frac{1}{2}at^2$ Where S is the distance, u is the initial velocity, t is time and a is acceleration.

8. Callum29

I remember I proved it once but can't remember the other formulae I used in the proof, it's just a substitution thing anyway