A toy rocket is launched vertically with an initial velocity of 128 feet/second. What is it's maximum height.

Question

h= -16t² + 128t

It is a 4 part question.

1. How long will it take to return to the ground

2. How long will it take to reach 112 ft.

3. What is it's maximum height

4 How long will it take to reach it's maximum height

I got that 112 ft will take 1 second, but I don't see how to get the other answers.

Won't the descending velocity be different?

Do I need to know the velocity of the pull of gravity?

Answer 1

1. hitting the ground means the height is 0......so -16t² + 128t = 0.......-16t(t-8)=0.....t=0 and t=8.

    So you launch it time 0 and it lands after 8 seconds.

 

2. So you want to fin when the height is 112....so -16t² + 128t = 112.......0=16t² -128t + 112.....0=t²-8t+7...

    (t-1)(t-7)=0.....so at t=1, the rocket is going up and is 112 and at t=7, the rocket is about to land and reaches 112.

 

3. The maximum height is finding the turning point which is on the axis of symmetry, which has the formula x=-b/2a.

     So the axis of symmetry has the equation x= -128/2(-16) or x = 4. Now find the height by plugging it in to the equation

    to get the turning point (4,256), which means that at 4 seconds, the height is 256 ft.

 

4. It takes 4 seconds to reach the maximum height, for the same reasoning as part 3.

 

Hope this helped.

Answer 2

Dear Deborah,

     The equation includes all the important information.  Did you graph it?  Just replace your "t" for an "X."  Then you will be able to graph it.  However, you may want to change some Y- max's and Y-min's, so that you can see it.  This can be done by going to the window function, second function. Since the lowest X value is zero, change X to a minimum of zero.  The highest Y value is at 257 feet, so make Y-max =270.  If you go back to the table, you can see that the highest value is 256 feet.  Use your arrow keys to move up or down while you are in the Window function. 

 

The toy hits 112ft twice, not once.  As it is moving towards its peak point (vertex), it reaches 112 ft. Then as it is moving away from the peak point, it moves downward. Check the graph and the table. Also, you can set the equation equal to 112 ft and factor it, for example:

 

112= -16t² +128t     Subtract 112 from both sides.

 

0= -16t² +128t -112   Simplify by  multiplying every term by -1.  Then divide each term by 8.

0=2t² -16t +14        Simplify by dividing every term by 2. 

 

0=t² -8t +7                Now, you have an easier equation to factor.

 

0=(t-1)(t-7)               Set each factor equal to zero.

 

0=(t-1)  or  (t-7)=0          t-1=0      t=1 second                             t-7=0       t=7    

 

 

#2.  It will take one second to reach 112 feet for the first time, but it will reach 112 feet again at 7 seconds.

 

 

 

#1  How long will it take to return to the ground?  (You are looking for the second t-intercept, where its height=0.)

Just set your equation to equal zero.

 

h=-16t²+128t=-0                Factor  by pulling out -16t from each term.

 

0= -16t(t-8)                    -16t=0      or       t-8=0    Now, just solve for "t."

 

It will take the toy rocket 8 seconds to hit the ground.

 

 

#3.  What is the maximum height? (You are looking for the vertex point.)  Here is a quick way to do it:

 

Note:  Here is the standard form for a quadratic equation:   ax² +bX +C=0.  The equation given is already in this form.

 

 

  0= -16t² +128t           a=-16     b=128       

 

Then, just use these equations to obtain the coordinate for the vertex:

 

h= - (b)/2a       and  k= f(h)           (h,k)      h=t,    and k= height of rocket  (maximum point)

 

h=-(128)/2(-16)= -128/-32  = 4       f(4)=  -16(4)² +128(4) =-16(16) +512=-256 +512=256  

 

The vertex is at (4, 256).  The maximum height= 256 feet.  You can look at the table on your graphing calculator.  The points after this point start to move toward the ground.             

 

I hope that I have helped you, if I have please give me thumbs up. I am available for on-line tutoring.

 

                                                                  Susan C.