SOLUTION: If a rock is thrown upward with an initial velocity of 32 feet per second from the top of a 40-story building,write the height equation using this information. How high is the roc

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Question 177100This question is from textbook Blitzer College Algebra
: If a rock is thrown upward with an initial velocity of 32 feet per second from the top of a 40-story building,write the height equation using this information.
How high is the rock after 0.5 seconds? How many seconds will the rock reach maximum height? What is the maximum height? This question is from textbook Blitzer College Algebra

Found 3 solutions by jim_thompson5910, amalm06, Alan3354:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
"write the height equation using this information"

Remember, the general height equation is s=-16t%5E2%2Bv%5B0%5D%2Bs%5B0%5D where "s" is the position of the object, "t" is the time, v%5B0%5D is the initial velocity, and s%5B0%5D is the initial position.


Since the "initial velocity of 32 feet per secound from the top of a 40 foot building", this means that v%5B0%5D=32 and s%5B0%5D=40


So the equation is s=-16t%5E2%2B32t%2B40 (after plugging in the initial velocity and position)


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b)

"how high is the rock after 5.0 seconds?"

Are you sure that the "5" isn't "0.5"? After 5 seconds, the rock wouldn't be in the air.



s=-16t%5E2%2B32t%2B40 Start with the given equation.


s=-16%280.5%29%5E2%2B32%280.5%29%2B40 Plug in t=0.5.


s=-16%280.25%29%2B32%280.5%29%2B40 Square 0.5 to get 0.25.


s=-4%2B32%280.5%29%2B40 Multiply -16 and 0.25 to get -4.


s=-4%2B16%2B40 Multiply 32 and 0.5 to get 16.


s=52 Combine like terms.


So after 0.5 seconds (half a second), the object is 52 ft in the air.


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c)


"after how many seconds will the rock reach maximum height"


To find the time where the object reaches the max height, we need to find the x-coordinate vertex of s=-16t%5E2%2B32t%2B40


t=%28-b%29%2F%282a%29 Start with the vertex formula.


From s=-16t%5E2%2B32t%2B40, we can see that a=-16, b=32, and c=40.


t=%28-%2832%29%29%2F%282%28-16%29%29 Plug in a=-16 and b=32.


t=%28-32%29%2F%28-32%29 Multiply 2 and -16 to get -32.


t=1 Divide.


So at one second, the object will reach the max height.


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d)

"what is the maximum height?"


s=-16t%5E2%2B32t%2B40 Start with the given equation.


s=-16%281%29%5E2%2B32%281%29%2B40 Plug in t=1 (the time at which the object will reach the peak).


s=-16%281%29%2B32%281%29%2B40 Square 1 to get 1.


s=-16%2B32%281%29%2B40 Multiply -16 and 1 to get -16.


s=-16%2B32%2B40 Multiply 32 and 1 to get 32.


s=56 Combine like terms.


So the max height is 56 feet.

Answer by amalm06(224) About Me  (Show Source):
You can put this solution on YOUR website!
The problem can be solved using momentum and energy considerations. The net force acting on the rock causes the velocity of the rock to change. This is given by the following impulse-momentum equation:

FΔt=m(vf-v0)

Since gravity is the only force that acts on the rock:
mgΔt=m(vf-v0)
gΔt=vf-v0, where g=-32 ft/s^2.
Then (-32)(Δt)=vf-v0
At maximum height, the velocity of the rock is zero. If you imagine the trajectory of the rock to follow a symmetrical parabolic path, then the slope of the horizontal tangent is zero at maximum height. But this slope is just the first derivative of the position, which is the velocity.
Then (-32)((Δt)=0-32 = -32
Therefore, t=1. The rock reaches maximum height after 1 second (Answer).

At maximum height, the rock has only potential energy. Since we are neglecting air resistance, friction does no work on the rock:
(1/2)(m)(V^2)i=mg(h2-40)
0.5(v^2)i=g(h2-40)
(0.5)(32)=(32)(h2-40)
16=h2-40
h2=56 ft. The maximum height of the rock is 56 ft (Answer)

To find the height of the rock after 0.5 seconds, use the following formula:
Δy=Vit + 0.5a(t^2)
So that y2-40=32t+(0.5)(-32)(t^2)
y2=32t-16(t^2)+40
y2=-16(t^2)+32t+40

For t=0.5 seconds, y2=-4+16+40=52 ft (Answer)




Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
If a rock is thrown upward with an initial velocity of 32 feet per second from the top of a 40-story building,write the height equation using this information.
How high is the rock after 0.5 seconds? How many seconds will the rock reach maximum height? What is the maximum height?
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What's the height of 40 stories?
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Using 400 ft for the 40 stories:
h(t) = -16t^2 + 32t + 400 --- h in feet, t in seconds
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How high is the rock after 0.5 seconds?
h(0.5) = -16*(0.25) + 32*0.5 + 400 = 412 feet
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How many seconds will the rock reach maximum height?
That's the vertex of the parabola at t = -b/2a
t = -32/-32 = 1 second.
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What is the maximum height?
h(1) = -16 + 32 + 400 = 416 feet.