SOLUTION: The path of a falling object is given by the function s=-16t^+v0t+s0 where the v0 represents the initial velocity in ft/sec ans s0represents the initial height. The variable t is t

Question 176994: The path of a falling object is given by the function s=-16t^+v0t+s0 where the v0 represents the initial velocity in ft/sec ans s0represents the initial height. The variable t is time in seconds, and s is the height of the object in feet. if a rock is thrown upward with the initial velocity of 32 feet per secound from the top of a 40 foot building a) write the height equation b)how high is the rock after 5.0 seconds,after how many seconds will the rock reach maximum height and what is the height?
I have been stuck on this problem for days I do not know where it is from it is a problem given by our teacher and I have no clue where it is from and that is the truth.
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
a)

Since the "initial velocity of 32 feet per secound from the top of a 40 foot building", this means that and

So the equation is (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.

Start with the given equation.

Plug in .

Square to get .

Multiply and to get .

Multiply and to get .

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

Start with the vertex formula.

From , we can see that , , and .

Plug in and .

Multiply 2 and to get .

Divide.

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

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d)

"what is the (max) height?"

Start with the given equation.

Plug in (the time at which the object will reach the peak).

Square to get .

Multiply and to get .

Multiply and to get .

Combine like terms.

So the max height is 56 feet.

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

The initial velocity is a given ("...thrown upward with the initial velocity of 32 feet per secound[sic]...")

The initial height is a given ("...from the top of a 40 foot building...")

So just just substitute:

Now, the question is what is the height at 5 seconds. is the height function with respect to time, so just substitute 5 for t and do the arithmetic.

which means that the rock is 200 feet below the ground. An absurd result to be sure. Double check the problem you were given because I suspect that the question was to determine the height at 0.5 seconds instead of the 5.0 seconds that you stated.

The second part of your problem needs you to recognize that the height function is a parabola and that it is concave down because the lead coefficient is negative. That means the vertex of the parabola is a maximum.

For any parabola of the form , the x-coordinate of the vertex is at .

For your problem, and , so the maximum height is reached at time second.

Calculate to get the actual maximum height reached.
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