Parabolas Application Q2

Question

The height, h metres, of a model rocket above the ground t seconds after after launching is given by the equation h=4t(50-t) for 0 < t < 50

 

a) Sketch the graph of the rocket when it is launched

b) Find the height of the rocket when it is launched

c) What is the greatest height the rocket reaches?

d) After how long does the rocket reach the greatest height?

How long is the rocket in the air?

 

I really need helps. Its parabola application and needs working out.

 

Answer 1

Hi There,

 

We'll consider each question in turn.

 

 

Part A (sketch the graph):

 

Below is the link to our graph of the parabola for the height of the rocket:

 

https://dl.dropbox.com/s/d2cq8s5v6g7jlvr/Graph_of_Parabola_for_Height_of_Rocket.png?raw=1

 

 

Part B (height of rocket at launch):

 

At launch (t=0), using our equation:

    h = 4t(50-t)

    h = 4(0)(50-0)

    h = (0)(50)

    h = 0 meters (height of rocket at launch)  

           [point (0,0) on graph - Part A]

 

 

Part C (greatest height of rocket):

 

The vertex (maximum point) can be determined by using the axis of symmetry formula to find time (t) and then plug that value into our equation to calculate the maximum height:

    h = 4t(50-t)

    h = -4t² + 200t  [a is -4 and b is 200)

    t = -b/2a  [axis of symmetry formula]

    t = -200 / 2(-4)

    t = -200/-8

    t = 25 seconds (time to reach greatest height)

    h = -4(25)² + 200(25)

    h = -4(625) + 5,000

    h = -2,500 + 5,000

    h = 2,500 meters (greatest height of rocket)  

          [point (25,2500) on graph - Part A]

 

 

Part D (time of rocket in air):

 

At launch and at landing (h=0), using our equation:

    h = 4t(50 - t)

    4t(50 - t) = 0

    -4t² + 200t = 0

    t² - 50t = 0

    t(t - 50) = 0

    t = 0 | t - 50 = 0

            | t = 50

    t = 50 seconds (time of rocket in air)

         [point (50,0) on graph - Part A]

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan.