The height in feet of a free falling object t seconds after release is s(t)=-16t^2+ v_0t+s_0, where s_0 is the height(in feet) at which the object is realsed, and v_0 is the initial velocity (in feet per second). Suppose the coin is dropped from a height of 1454 feet.

Question

D. At what time is the instanteous velocity equal to the average velocity if the coin found in part B?
Part B answer is -64ft/ sec
I do not know where to begin here please help

Answer 1

v = ds/dt
=-16(2)t
=-32 t

-32 t = -64 ??
if that is the question the answer is 2

Answer 2

Did you not look at my solution to the same question ?
http://www.jiskha.com/display.cgi?id=1468890453

All I can see wrong with it, I labeled part D as B

for B all you have to do it sub t = 1 and t = 3 into your velocity equation:
v = -32t.