SOLUTION: If an object is propelled upward from ground level with an intial velocity of 64 ft. per sec, its height h in feet t seconds later is:
h=-16t^2 = 64t
(a) After how many seconds i
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h=-16t^2 = 64t
(a) After how many seconds i
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Question 336792: If an object is propelled upward from ground level with an intial velocity of 64 ft. per sec, its height h in feet t seconds later is:
h=-16t^2 = 64t
(a) After how many seconds is the height 48 ft?
(b) The object reaches its maximum height 2 sec after it is propelled. What is this maximum height?
(c) After how many seconds does the object hit the ground?
I have read all of the examples that they give me in my math book, but i still have a hard time trying to understand these problems. Found 2 solutions by stanbon, nyc_function:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! If an object is propelled upward from ground level with an intial velocity of 64 ft. per sec, its height h in feet t seconds later is:
h=-16t^2 + 64t
(a) After how many seconds is the height 48 ft?
Let h=48 and solve for "t":
-16t^2+64t-48 = 0
-16(t^2-4t+3) = 0
(t-3)(t-1) = -
t= 1 sec or t = 3 sec
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Ans: time = 1 sec on the way up and time = 3 sec on the way down.
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(b) The object reaches its maximum height 2 sec after it is propelled. What is this maximum height?
f(x) = -16t^2+64t
f(2) = -16(4)+64(2) = 64 ft.
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(c) After how many seconds does the object hit the ground?
Let h = 0 and solve for "t":
-16t^2+64t = 0
-16t(t-4) = 0
t = 0 or t = 4 seconds
t = 0 seconds refers to the time at which the object is launched.
t = 4 seconds refers to the time at which the object is back on the ground.
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Cheers,
Stan H.
You can put this solution on YOUR website! Actually, I think you meant
h = -16t^2 + 64t since initial velocity is upward.
a] 48 = -16t^2 + 64t... use the quadratic formula
b] h(2)
c] -16t^2 + 64t = 0 will have two roots, one for t=0 or launch, and the other when it hits the ground.
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