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Question 585157: A bullet is shot vertically upward to a velocity of 96 ft/sec. The height h is in feet, atter t seconds is expressed by h=96t -16t^2. At what times was the bullet on the ground? At what time did the bullet reach its max height? What was the max height?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! A bullet is shot vertically upward to a velocity of 96 ft/sec. The height h is in feet, atter t seconds is expressed by h=96t -16t^2. At what times was the bullet on the ground? At what time did the bullet reach its max height? What was the max height?
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h=96t -16t^2
solving for t when bullet on the ground.
h=0
0=96t -16t^2
complete the square
-16t(t-6)=0
t=0 (reject, t>0)
t-6=0
t=6 seconds
..
h=96t -16t^2
complete the square
h=-16(t^2-6t+9)+144
h=-16(t-3)^2+144
This equation is that of a parabola that opens downwards with vertex at (3,144)
..
ans:
Bullet hit the ground in 6 sec
Bullet reaches maximum height in 3 sec
Maximum height=144 feet
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