SOLUTION: h=-16t^2+64t A. The object reaches its maximum height 2 seconds after it is propelled. What is the maximum height? B. After how many seconds does the object hit the ground?

Question 432312: h=-16t^2+64t
A. The object reaches its maximum height 2 seconds after it is propelled. What is the maximum height?
B. After how many seconds does the object hit the ground?

Please help me work this out!!
Thanks,
Chris
Found 2 solutions by stanbon, shree840:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
h=-16t^2+64t
A. The object reaches its maximum height 2 seconds after it is propelled. What is the maximum height?
h(2) = -16(2)^2+64(2) = -16*4 + 128 = -64+128 = +64 ft.
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B. After how many seconds does the object hit the ground?
Solve -16t^2+64t = 0
-16t(t-4) = 0
Positive solution:
t = 4 seconds
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Cheers,
Stan H.
Answer by shree840(260) (Show Source): You can put this solution on YOUR website!
your formula is given h=-16T^2+64t
substitute t=2
h=-16(2)^2+64*2
=-64+128=64 ANS
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