SOLUTION: Everyone in my family has tried to answer this worksheet from school. Does it need to be solved before graphing? Would appreciate your help with this problem. Thanks! A roc

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Question 81891: Everyone in my family has tried to answer this worksheet from school. Does it need to be solved before graphing? Would appreciate your help with this problem. Thanks!
A rocket is launched from atop a 78-foot cliff with an initial velocity of 152 feet per second. the height of the rocket t seconds after launch is given by the equation h=-16t^2 + 152t + 78. Graph the equation to find out how long after the rocket is launched it will hit the ground. Estimate the answer to the nearest tenth of a second.
Found 2 solutions by Earlsdon, bucky:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The instruction do indicate to determine the solution by graphing, so let's see what the graph looks like:
Graph the function:h%28t%29+=+-16t%5E2%2B152t%2B78
Instead of graphing x (horizontally) and y (vertically) as in the usual graph, here, we'll be graphing t (horizontally) and h (veritically).
In the resulting graph, we'll be looking for where the graph crosses the t-axis because this is when the height of the rocket will be zero (ground-level).
graph%28300%2C200%2C-5%2C15%2C-25%2C450%2C-16x%5E2%2B152x%2B78%29
Looking at positive t-axis only (negative time is not meaningful here) the graph indicates that the curve (parabola) crosses the t-axis at about t = 10 seconds.
The rocket will hill the ground about 10 seconds after launch.
Let's check this solution algebraically:
h%28t%29+=+-16t%5E2%2B152t%2B78
We need to find the "zeros" or the roots of this quadratic equation.
Using the quadratic formula:x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
Of course, we are looking for t rather than x, so...
t+=+%28-152%2B-sqrt%28152%5E2-4%28-16%29%2878%29%29%29%2F2%28-16%29
t+=+%28-152%2B-sqrt%2828096%29%29%2F-32
The roots are:
t+=+9.988 and
t+=+-0.488 Discard this solution as a negative time is not meaningful.

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
h+=+-16t%5E2+%2B+152t+%2B+78
.
You can solve this problem by graphing only. Label the vertical axis (normally the y-axis)
as height and label the horizontal axis (normally the x-axis) as time. Then you can make
the graph by picking values of t, substituting those values into the equation and finding
the corresponding value of h, the height.
.
For example, when t is zero the height is:
.
h+=+-16%280%5E2%29+%2B+152%280%29+%2B+78+=+0+%2B+0+%2B+78+=+78
.
So when the time is zero, the height is 78 feet, but the problem told you that.
.
On the height (vertical axis) put a dot at 78 feet.
.
Next try letting t = 1 second. The equation then becomes:
.
ft
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Plot the point (1, 214). That's a second point on the graph. If you continue picking more
values for t, you will begin to see a pattern in which the object rises to a peak, and then
starts to drop back to Earth.
.
It takes a lot of work to build a graph that is complete enough to get the time it hits
the ground down to 0.1 second accuracy.
.
So let's do it mathematically. It will tell you where to concentrate your graph.
.
Begin by recognizing that when the rocket hits the ground its height will be zero. So
in the equation, we can set h equal to zero and solve for t.
.
0+=+-16t%5E2+%2B+152t+%2B+78
.
Normally by convention we write this with the sides switched. So ...
.
-16t%5E2+%2B+152t+%2B+78+=+0
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This is of the standard form:
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at%5E2+%2B+b%2At+%2B+c+=+0
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and by comparison with our equation we can see that a = -16, b = 152, and c = 78.
.
The quadratic formula tells us that for equations in the standard form, the solutions
to the equation are given by the equation:
.
t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
.
So now all you have to do is substitute the values of a, b, and c and get:
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t+=+%28-152+%2B-+sqrt%28152%5E2+-+4%2A%28-16%29%2A78%29%29%2F%282%2A%28-16%29%29
.
Now it's just a lot of calculator work. Note that 152^2 = 23104 and -4*(-16)*78 = 4992. Substitute
and get:
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t+=+%28-152+%2B-+sqrt%2823104+%2B+4992%29%29%2F%28-32%29+
.
t+=+%28-152+%2B-+sqrt%2828096%29%29%2F%28-32%29+=+%28-152+%2B-+167.6186147%29%2F%28-32%29
.
This will lead to two answers ... one with the plus sign and one with the minus sign. Let's solve
for the two possible answers. First solve using the plus sign. That leads to:
.
t+=+%28-152+%2B+167.6186147%29%2F%28-32%29+=+15.61861472%2F%28-32%29+=+-+0.488081709
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But this cannot be ... a minus time means before launch. So we know this answer is wrong.
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So let's solve for the other possible answer. This time use the minus sign:
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t+=+%28-152+-+167.6186147%29%2F%28-32%29+=+-319.6186147%2F%28-32%29+=+9.98808171
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This tells us that 9.98808171 seconds after launch the rocket hits the ground. Now you
know what region of time you need to concentrate on. To the nearest tenth of a second the
answer is 10.0 seconds).
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And for further information, the highest the rocket will get in its flight occurs when
the time is -b%2F%282%2Aa%29 seconds. Substituting the values of "b" and "a" above the
answer becomes:
.
-152%2F%282%2A%28-16%29%29+=+-152%2F-32+=+4.75
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When t = 4.75 seconds after launch, the rocket will be at its maximum height. All you
have to do to find the maximum height is to use the height equation and set t equal to
4.75 seconds:
.

.
So in your graphing you know that the maximum height the rocket will reach is 439 feet
and that will occur when the time is 4.75 seconds after launch.
.
Hope this helps you to understand and do the problem.