SOLUTION: Solve: If an object is thrown upward with an initial velocity of 64ft/sec, its height after t sec is given by h=64t-16t^2. Find the number of seconds the object is in the air be

Question 145012: Solve:
If an object is thrown upward with an initial velocity of 64ft/sec, its height after t sec is given by h=64t-16t^2. Find the number of seconds the object is in the air before it hits the ground.
Any assistance you can provide me with is greatly appreciated!
Found 2 solutions by stanbon, shahid:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
If an object is thrown upward with an initial velocity of 64ft/sec, its height after t sec is given by h=64t-16t^2. Find the number of seconds the object is in the air before it hits the ground.
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h(t) = 64t-16t^2
Height is zero when the object hits the ground.
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64t - 16t^2 = 0
-16t(-4 + t) = 0
t = 0 or to = 4 seconds
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The object is in the air 4 seconds.
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Cheers,
Stan H.
Answer by shahid(44) (Show Source): You can put this solution on YOUR website!
if height of the object is given as a function of time so its derivative gives velocity.as ball is thrown upward so its final velocity will be zero or dh/dt=0
differentiate h=64t-16t^2 w.r.t
dh/dt= 64-32t put dh/dt=0
64-32t=0
t=2 seconds.ball takes 2 seconds to attend maximum height and 2 seconds to return so total 4 seconds it remains in the air
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