sarahmd0102
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A rocket is launched from a 50 foot platform with an initial velocity of 300ft/sec.

1) create a function h(t) for the path of the rocket.

2) when will the rocket reach its max height?

3) what is the rocket’s max height?

4) when will the rocket land on the ground?

5) how long was the rocket above 1200 feet?



Answer :

Answer:

1) h(t) = -16t² + 300t + 50

2) h'(t) = -32t + 300

-32t + 300 = 0

t = 9.375 sec.

3) h(9.375) = 1,456.25 ft.

4) 16t² - 300t - 50 = 0

8t² - 150t - 25 = 0

(-150)² - 4(8)(-25) = 23,300

t = (150 ± √23,300)/16

= (150 + 10√233)/16

= (75 + 5√233)/8

= about 18.92 sec.

5) -16t² + 300t + 50 = 1,200

-16t² + 300t - 1,150 = 0

16t² - 300t + 1,150 = 0

8t² - 150t + 525 = 0

(-150)² - 4(8)(525) = 5,700

t = (150 ± √5,700)/16

= (150 ± 10√57)/16

= (75 ± 5√57)/8

4.66 sec. < t < 14.09 sec.

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