s(t) = 180-5t^2
(a) avg velocity is total distance/total time.
s(2) = 180-5*4 = 160
160m/2s = 80 m/s
v(t) = ds/dt = -10t
v(4) = -40
180-5t^2 = 0
t = 6
That is, at t=6, the height is zero -- it hit the ground.
Now you just plug 6 into v(t)
The height, in metres, of an object that has fallen from a height of 180 m is given by the position function s(t)=-5t^2 +180, where t >0 and t is in seconds.
a. Find the average velocity during each of the first two seconds.
b. Find the velocity of the object when t =4.
c. At what velocity will the object hit the ground?
1 answer
1 answer