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

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acceleration due to gravity=9.8 m/s^2
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: \[S= ut + at^2 = 0 + 9.8(4)^2 = 156.8m\]
I understand that but I'm not sure I set the equation up right. I used: \[DeltaX=1/2at ^{2}\]

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sorry, \[S= ut + \frac{1}{2}at^2 = 4.9\times 16 = 78.4m\]
Okay so the 1/2 is right?
that is the formula one should use...
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.
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

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