NASA launches a rocket at t = 0 seconds. Its height, in meters above sea level, as a function of time is given by h(t) = – 4.9t2 + 139t + 185,
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
The rocket splashes down after
seconds.
How high above sea level does the rocket get at its peak?
The rocket peaks at
meters above sea level.

Step-by-step explanation:

y = ax² + bx + c

D = b² - 4ac

= ( - b ± √D ) ÷ 2a

= -

~~~~~~~~~~~~~

h(t) = - 4.9t² + 139t + 185

- 4.9t² + 139t + 185 = 0

D = 139² + 4(- 4.9)(185) = 22947

= ( - 139 ± √22947) ÷ 2(- 4.9)

≈ 29.641 ≈ 30 seconds

< 0

= - 4.9( )² + 139( ) + 185 ≈ 1170.7653 ≈ 1171 meters

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step-by-step explanation:

y = i/2 xtnop + (3i/2 tnop+ 1)

step-by-step explanation:

the picture is the step by step explanation.