d = V(0) + 0.5at^2
d = 0 + 0.5(32.2)(4)^2
d = 257.6 feet

http://en.wikipedia.org/wiki/Equations_for_a_falling_body

d= vt + .5at^2
d= 0 + .5*9.8(4)^2
d= .5*9.8*16
d= 78.4m
d= 257 ft

so it's C)

i get 156meters ... so idfk whut the answer is kuz thats over 300 feet

v = a * t

v = 9.8 * 4

average velocity = 39.2 m/s

v*t = D

39.2 * 4 = 156 meters




ahhh yea forgot about that equation.. hes right

This post was edited by Tboner on Jun 6 2011 09:23pm

@tboner
your first equation is wrong it is
V average = (V max + V min) / 2

at time = 0 it is going 0 m/s
at time = 1 it is going 9.81 m/s
at time = 2 it is going 19.62 m/s
and so on (these are estimates)

thats where your getting such a high number from

Jct and Exx have the right solution, in metric it works out as follows:

the full formaula is :
change in d = initial velocity * change in time + .5 * acceleration * change in time ^2
d = 0 +(.5)*(9.81 m/s)(4)^2
d = 78.48 meters which equals the 257 feet