SOLUTION: 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 si
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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?
b.At what time was the football less than its height when punted?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
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
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