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catsrule332
 4 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.
catsrule332
 4 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.

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loophole144
 4 years ago
Best ResponseYou've already chosen the best response.0acceleration due to gravity=9.8 m/s^2

Callum29
 4 years ago
Best ResponseYou've already chosen the best response.2If 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: \[S= ut + at^2 = 0 + 9.8(4)^2 = 156.8m\]

catsrule332
 4 years ago
Best ResponseYou've already chosen the best response.0I understand that but I'm not sure I set the equation up right. I used: \[DeltaX=1/2at ^{2}\]

Callum29
 4 years ago
Best ResponseYou've already chosen the best response.2sorry, \[S= ut + \frac{1}{2}at^2 = 4.9\times 16 = 78.4m\]

catsrule332
 4 years ago
Best ResponseYou've already chosen the best response.0Okay so the 1/2 is right?

Callum29
 4 years ago
Best ResponseYou've already chosen the best response.2that is the formula one should use...

Callum29
 4 years ago
Best ResponseYou've already chosen the best response.2Yeah, 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.

Callum29
 4 years ago
Best ResponseYou've already chosen the best response.2I 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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