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A.T:

--- Quote from: astarmathsandphysics on March 24, 2010, 02:08:48 pm ---Because 330 is in the range 327.66 -3% to 327.66+3%

--- End quote ---
Arent all the values in that range?
i dont think this is the correct logic.

ruby92:
an arrow is shot horizontally at a target 20m away.The arrow hits a point 19.6cm verically below the point it was aimed at.Calculate the speed with which the arrow left the bow.


could someone please explain this and the above questions...urgently!!!! ???

astarmathsandphysics:
y=1/2at^2
0.196=4.9t^2 so t=0.2 seconmd
v=x/t=20/0.2 =100m/s

ruby92:
hw did u get 4.9 as the acceleration

A.T:

--- Quote from: ruby92 on March 24, 2010, 09:48:21 pm ---hw did u get 4.9 as the acceleration

--- End quote ---
9.81/2
9.81 = g

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