SOLUTION: A fireworks shell is shot straight up with an initial velocity of 80 feet per second. Its height s after t seconds is given by the equation
s = 80t − 16t2.
If the shell i
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Question 908390: A fireworks shell is shot straight up with an initial velocity of 80 feet per second. Its height s after t seconds is given by the equation
s = 80t − 16t2.
If the shell is designed to explode when it reaches its maximum height, how long after being fired, and at what height, will the fireworks appear in the sky?
time=_______ sec
height=_________ft
Found 3 solutions by lwsshak3, Alan3354, stanbon:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
A fireworks shell is shot straight up with an initial velocity of 80 feet per second. Its height s after t seconds is given by the equation
s = 80t − 16t2.
If the shell is designed to explode when it reaches its maximum height, how long after being fired, and at what height, will the fireworks appear in the sky?
***
s=80t-16t^2
s=-16t^2+80t
complete the square
s=-16(t^2-5t+25/4)+100
s=-16(t-2.5)^2+100
how long after being fired, and at what height, will the fireworks appear in the sky? 2.5 sec at a height of 100 ft
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A fireworks shell is shot straight up with an initial velocity of 80 feet per second. Its height s after t seconds is given by the equation
s = 80t − 16t2.
If the shell is designed to explode when it reaches its maximum height, how long after being fired, and at what height, will the fireworks appear in the sky?
Don't burden us with gibberish.
s = 80t − 16t2. What is that?
-----
h(t) = 80t - 16t^2
Use ^ for exponents, Shift 6
----------------------
The max ht is the vertex of the parabola.
The vertex is at t = -b/2a = 80/32 = 2.5 seconds
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t = 2.5 seconds
h(2.5) = 200 - 100 = 100 feet
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don't waste your time and mine with this
time=_______ sec
height=_________ft
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Its height s after t seconds is given by the equation s = 80t − 16t^2.
If the shell is designed to explode when it reaches its maximum height, how long after being fired?
-----
Max occurs when x = -b/(2a) = -80/(2*-16) = 5/2 seconds
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And at what height, will the fireworks appear in the sky?
f(5/2) = 80(5/2) - 16(5/2)^2 = 200 - 20 = 180 ft
============================
Cheers,
Stan H.
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