## catsrule332 Group Title A rock is pushed off of a cliff. If air resistance can be neglected, how far will the rock have fallen after 4s? Explain. 2 years ago 2 years ago

1. loophole144 Group Title

acceleration due to gravity=9.8 m/s^2

2. Callum29 Group Title

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 Group Title

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

4. Callum29 Group Title

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

5. catsrule332 Group Title

Okay so the 1/2 is right?

6. Callum29 Group Title

that is the formula one should use...

7. Callum29 Group Title

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 Group Title

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