Question 666571
One day at football practice, Darrell, the kicker, punted the ball so that its height in feet above the ground was given by h(t)=-16t^2+40t+4
where t is the number of seconds since the ball was punted.
a.At what time was the football 20 feet or higher above the ground?
.
set h(t) to 20 and solve for t:
h(t)=-16t^2+40t+4
20=-16t^2+40t+4
0=-16t^2+40t-16
divide through by -8:
0=2t^2-5t+2
factoring:
0 = (2t-1)(t-2)
t = {1/2, 2}
so, answer is:
when t is between .5 seconds and 2 seconds
.
b.At what time was the football less than its height when punted?
from:
h(t)=-16t^2+40t+4
we know the "initial" height was 4 feet:
set h(t) to 4 and solve for t:
h(t)=-16t^2+40t+4
4=-16t^2+40t+4
0=-16t^2+40t
0=16t^2-40t
0=2t^2-5t
0=t(2t-5)
t = {0, 5/2}
so, the answer is 
when t is "less than" 5/2 seconds or 2.5 seconds