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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
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

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

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

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

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

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

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

Callum29 Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

Callum29 Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago
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