A missile is fired such that its height above the ground is given by h=-9.8t^2+38.2t+6.5, where t represents the number of seconds since the rocket was fired. Using the quadratic formula, determine, to the nearest tenth of a second, when the rocket will hit the ground.


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Respuesta :

Answer:

4.1 sec

Step-by-step explanation:

-9.8t^2 +38.2t + 6.5 = 0

t = [tex]\frac{-38.2 +-\sqrt{38.2^2{ - 4(-9.8)(6.5)} } }{2(9.8}[/tex]

 =  [tex]\frac{-38.2 +- \sqrt{1459.24 + 254.8} }{-19.6}[/tex]

 = [tex]\frac{-38.2 +- \sqrt{1714.04} }{-19.6}[/tex]

 = [tex]\frac{-38.2 +- 41.4}{-19.6}[/tex]

 = [tex]\frac{-79.6}{-19.6}[/tex]

 = 4.1

The rocket will hit the ground at 0.2 seconds

Function

The function is given as:

[tex]h=-9.8t^2+38.2t+6.5[/tex]

When the missile hits the ground, the value of height is 0

This means that:

[tex]h = 0[/tex]

So, we have:

[tex]-9.8t^2+38.2t+6.5 = 0[/tex]

Graph

Next, we plot the graph of [tex]-9.8t^2+38.2t+6.5 = 0[/tex]

See attachment for the graph of [tex]-9.8t^2+38.2t+6.5 = 0[/tex]

From the attached graph, we have:

[tex]t = (0.163, -4.061)[/tex]

Approximate

[tex]t = (0.2, -4.1)[/tex]

Time cannot be negative.

So, we have:

[tex]t = 0.2[/tex]

Hence, the rocket will hit the ground at 0.2 seconds

Read more about height functions at:

https://brainly.com/question/10837575

Ver imagen MrRoyal