Question 475836: An arrow is fired straight up from the ground with an initial velocity of176 feet per second. Its height ,s(t), in feet at any time is given by the function s(t)=-16t^2+176t. Find the interval of time for which the height of the arrow is greater than 160 feet.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! An arrow is fired straight up from the ground with an initial velocity of176 feet per second. Its height ,s(t), in feet at any time is given by the function s(t)=-16t^2+176t. Find the interval of time for which the height of the arrow is greater than 160 feet.
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calculating time at which arrow reaches 160 ft
s(t)=-16t^2+176t
160=-16t^2+176t
0=-16t^2+176t-160
0=16t^2-176t+160
0=16t^2-176t+160
Divide by 16
0=t^2-11t+10
0=(t-1)(t-10)
t=1 sec (time at which arrow reaches 160 ft)
t=10 sec (reject, exceeds time arrow reaches maximum height)
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Calculating time at which arrow reaches maximum height
y=-16t^2+176t
completing the square
y=-16(t^2-11+121/4)+484
y=-16(t-11/2)^2+484
arrow reaches maximum height at 11/2 or 5.5 sec
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Arrow takes 1 sec to reach 160 ft on the way up, plus 4.5 more sec to reach maximum and 4.5 more sec to descend to 160 ft on the way down.
So, interval of time arrow is above 160 ft is: 1 < t < 10, that is, between 1 and 10 seconds
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