SOLUTION: Thank you so much for the answer earlier I really apprecate it. I have one final question of the day hope you can help me out. Thanks again if a rock is thrown upward with an i

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Question 112941: Thank you so much for the answer earlier I really apprecate it. I have one final question of the day hope you can help me out. Thanks again
if a rock is thrown upward with an initial velocity of 64 feet per second from the top of a 25 foot building write the a) height(s) equation using this information typing hint:t^2
b) how high is the rock after 1 second, show work
c)after how many seconds will the graph reach maximum height, show work
d)what is maximum height, show work

If you can help me with this it will save my life thank you again.
Found 2 solutions by bucky, solver91311:
Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website!
The equation for the height of a "free falling body in motion" is:
.

.
where:
is the height of the object at a time equal to or greater than zero;
is the time at or after release of the object
is the initial velocity of the object at launch (+ is up and - is down}
is the height from which the object is launched
.
The problem tells you that and . Substitute these values and
the height equation becomes:
.

.
and this is the answer to part a of your problem.
.
part b asks how high the object is after 1 second. Just plug 1 in for t and you get:
.

.
So after 1 second the object is 73 feet above the ground.
.
part c asks at what time the object will reach its maximum height. This can be found by
recognizing that the height equation is a quadratic equation of the standard form:
.

.
if the quadratic formula is applied to solve this equation, the vertex of the path will occur
where
.
By comparing terms of the equation that is the answer for part a with the terms of the
standard form of the quadratic equation, you find that a = -16 and b = +64. Therefore,
you can get that the vertex occurs at a time equal to:
.

.
So the peak height occurs at t = 2 seconds after launch of the object. This is the answer
to part c.
.
part d asks how high the object gets at its peak. We have found that the answer occurs
when t is 2 seconds. So all that we have to do to get the peak height is to return to
the equation we got for part a and substitute 2 for 2. This is done as follows:
.
Start with:
.

.
Substitute 2 for t and you get:
.

.
The answer is that the object reaches a maximum height above ground of 89 feet.
.
Hope this helps you to understand the problem.
.

Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!
The basic formula to use is where s(t) is the distance or height function with respect to time, is the initial or starting height, is the initial velocity, is the constant acceleration due to gravity, and is the time.

The first thing to realize is that . Why is it negative? Because the gravity vector is downward and you are throwing the rock upwards.

The other constants are given:

. You could make an argument that the initial height is 25 feet plus the distance from the roof of the building to the end of your outstretched arm at the moment you release the rock, but I think for the purposes of this problem you can ignore all of that.

The first part of the problem asks how high the rock will be after 1 second. Simply substitute 1 for t and evaluate:

So after 1 second, the rock will be at 73 feet above the ground.

Second part of the problem:

The graph of is a parabola that is convex down (the coefficient on the term is < 0). If we can find the t value of the vertex, we will have the time when the rock reaches maximum height.

The vertex of is given by . So for our problem, that is . So maximum height will be reached in 2 seconds.

Third part:
The maximum height is given by evaluating the function at the time we just determined in the previous part of the problem:

So the maximum height is 89 feet.

Hope that helps.
John