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djverab [1.8K]
3 years ago
13

A model rocket is launched from the ground with an initial velocity of 160 ft/sec. how long will it take the rocket to reach its

maximum height? Show all work in the space provided. Assume the model rocket’s parachute failed to deploy and the rocket fell back to the ground. How long would it take the rocket to return to Earth from the time it was launched? Show all work in the space provided.
Mathematics
1 answer:
timama [110]3 years ago
8 0
The formula we need is
h(t)=-16t^2+v_0 t+h_0, where <em>v</em>₀ is the starting velocity and <em>h</em>₀ is the initial height.  Using the velocity and starting height from our problem we have
h(t)=-16t^2+160t+0.  The path of this rocket will be a downward facing parabola, so there will be a maximum.  This maximum will be at the vertex of the graph.  To find the vertex we start out with x= \frac{-b}{2a}, which in our case is x= \frac{-160}{2(-16)}= \frac{-160}{-32} = 5.  It will take 5 seconds for the rocket to reach its maximum height.  Plugging this back into our formula gives us
h(5)=-16(5^2)+160(5)+0&#10;\\=-16(25)+800&#10;\\=-400+800=400
The rocket's maximum height is 400 feet.
We set our formula equal to zero to find the time it takes to hit the ground, then we factor:
0=-16t^2+160t+0&#10;\\0=-16t^2+160t&#10;\\0=-16t(t-10)
Using the zero product property, we know that either -16t =0 or t-10=0.  When -16t=0 is at t=0, when the rocket is launched. t-10=0 gives us an answer of t=10, so the rocket reaches the ground again at 10 seconds.
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Answer:

Kayla's reasoning is not correct.

Step-by-step explanation:

The locker problem is as follows:

Imagine you are at a school that has student lockers. There are 200 lockers, all shut and unlocked, and 200 students. Suppose the first student goes along the row and opens every locker. The second student then goes along and shuts every other locker beginning with number 2. The third student changes the state of every third locker beginning with number 3. (If the locker is open the student shuts it, and if the locker is closed the student opens it). The fourth student changes the state of every fourth locker beginning with number 4. Imagine that this continues until the 200 students have followed the pattern with the 200 lockers. At the end, which lockers will be open and which will be closed? Why?

Solution:

So from the information we know that the first student goes along the row and opens every locker.

Then the second student shuts every other locker, i.e. locker numbers 2, 4, 6, 8, 10, ..., 196, 198 and 200.

Then the third students changes the state of every third locker, i.e. he/she closes an open locker and opens a closed locker.

So the open lockers are: 1, 5, 6, 12,...

Then the fourth students changes the state of every fourth locker.

So the open lockers are: 1, 4, 5, 6, 8,....

So, on we will observe that the open lockers have a perfect square number such as, 1, 4, 9, 16,....

Consider that the pattern is as follows:

Student 1 opens the locker, Student 2 closes it, Student 3 opens it, person 4 Student and so on.

This is because the square numbers always have an odd number of factors, which leads them to be open at the end.

Take any locker number, 40, for example. Its state (open or closed) is changed for every student whose number in line is a factor of the locker number.

Student      Locker 40 status

     1                    Open

     2                   Close

     4                   Open

     5                   Close

     8                   Open

     10                  Close

    20                  Open

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Like all other lockers numbered with non-square numbers, it ends up closed after all the students have gone through the line because it has an even number of factors.

Consider the locker number 16:

Student      Locker 16 status

     1                    Open

     2                   Close

     4                   Open

     8                   Close

     16                  Open

Thus, we can conclude that all the doors with square numbers on them will remain open because all square numbers have an odd number of factors and the doors with non-square numbers on them will remain close because they have even number of factors.

There will be a total of 14 lockers open.

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169 and 196

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Answer:

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