SOLUTION: The formula h=-16t^2+vt gives the height h in feet of an object after t second, when it is shot upward into the air with an initial velocity v in feet per second. After how many se
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-> SOLUTION: The formula h=-16t^2+vt gives the height h in feet of an object after t second, when it is shot upward into the air with an initial velocity v in feet per second. After how many se
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Question 191039: The formula h=-16t^2+vt gives the height h in feet of an object after t second, when it is shot upward into the air with an initial velocity v in feet per second. After how many seconds will the object hit the ground if it is shot with a velocity of 144 feet per second? Found 2 solutions by nerdybill, stanbon:Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! Given:
h=-16t^2+vt
and
v = 144 feet per second
.
Set h=0 and solve for t:
h=-16t^2+vt
0=-16t^2+144t
0=-16t(t-9)
.
t = {0, 9}
answer: 9 seconds
You can put this solution on YOUR website! The formula h=-16t^2+vt gives the height h in feet of an object after t second, when it is shot upward into the air with an initial velocity v in feet per second. After how many seconds will the object hit the ground if it is shot with a velocity of 144 feet per second?
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h(t) = -16t^2 + vt
h(t) = -16t^2 + 144t
When the object hits the ground its height is zero.
Solve for t:
-16t^2 + 144t = 0
-16t(t - 9) = 0
t = 0 or to = 9
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The 0 means the object started on the ground.
The 9 means the object will again be on the ground after 9 seconds.
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Cheers,
Stan H.
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