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.
Please I need help with this question
Answers
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
FunctionThe 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]
GraphNext, 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:
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