Asked by Lorrie
A model rocket is launched from a roof of a building. Its path is described by the relationship h=-5t*2+35t+10, where h is the height in metres and t is the time in seconds. How tall is the building? What is the maximum height that the rocket will reach? When will the rocket be 40 metres above the ground? Explain your answer.
Answers
Answered by
Lorrie
The building is 10 metres tall.
h=-5t^2+35t+10
=-5(t^2-7t-2)
=-5(t^2-7t+12.25-12.25)+10
=-5(t^2-7t+12.25)+61.25+10
= -5(t^2-7t+12.25)+71.25
=-5(t-3.5)^2 + 71.25
So at 3.5 seconds the maximum height is 71.25 metres high.
Is this correct?
h=-5t^2+35t+10
=-5(t^2-7t-2)
=-5(t^2-7t+12.25-12.25)+10
=-5(t^2-7t+12.25)+61.25+10
= -5(t^2-7t+12.25)+71.25
=-5(t-3.5)^2 + 71.25
So at 3.5 seconds the maximum height is 71.25 metres high.
Is this correct?
Answered by
Lorrie
Are the first 2 parts correct?
Hint please on the third part? I think I need the velocity? How to calculate?
Hint please on the third part? I think I need the velocity? How to calculate?
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