s(t)=-16t^2+v(subscript0)t+(subscrpt0)

Question

a silver dollar is dropped from the top of a building that is 1362 feet tall. a) determine the position and velocity f'cn for the coin. b) determine the avg velocity on the interval [1,2] c) find the instantanous velocities when t=1 and t=2 d) find the time required for the coin to reach ground level. e) find the velocity of the coin at impact.

Answer 1

Good grief, feet and inches and yards. That is an old textbook with g=-32ft/second^2 so g/2 = 16
anyway
s = (1/2)gt^2 +Vo t + Ho
so
a) s = -16t^2 + 0 + 1362
ds/dt = velocity = -32 t + 0

and of course acceleration =d^2s/dt^2 = -32

b) find out how far it moved in that one second and divide by one second
s(2) = -16*4 +1362
s(1) = -16*1 +1362
difference = -48 ft in one second so average speed = -48 ft/second beteen t = 1 and t = 2

c) v(1) =-32(1) = -32 ft/s

v(2) = -32(2) = -64 ft/s

d) when will s = 0?
0 = -16t^2 + 0 + 1362
16 t^2 = 1362
t^2 = 681
t = 26.1 seconds

e) -16(26.1) =

Answer 2

-32 (26.1)

Answer 3

D) would actually be approximately 9.23 seconds. 1362/16=85.125

Answer 4

YOUR BOTH WRONG