Question

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)= 48t-16t^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?

Answer 1

a) evaluate using t=1
b) solve for t when s=0
c) using t from (b) evaluate s'(t)

Answer 2

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