calculus
posted by moises on .
if a ball is thrown vertically upward with an initial velocity of 48 ft/s, then its height, s, after t seconds is given by s(t)= 48t16t^2.
a) what is the velocity of the ball after 1 second?
b) when does the ball hit the ground?
c) with what velocity does the ball hit the ground?

a) evaluate using t=1
b) solve for t when s=0
c) using t from (b) evaluate s'(t) 
since t = 2,
i) at 0.5 seconds,
y= (52(2.5)16(2)^2)/0.5=13.6
ii) (52(2.1)16(2)^2)/0.1=12.8
and just add the time to 2 and divide by time