A rocket is fired vertically upward. Its height h(t) in meters above the ground at t seconds is given
by h = -4.9t² +232t + 185.
How high was the rocket when it was initially launched?

meters
How high is the rocket after 9 seconds?

meters
What is the velocity of the rocket after 9 seconds?

m/s
What is the acceleration of the rocket after 9 seconds?

m/s^2
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The rocket is at a height of 1876.1 meters after 9 seconds,the velocity of the rocket after 9 seconds is 143.8 m/s and  the acceleration of the rocket after 9 seconds is -9.8 m/s².

To find the height of the rocket when it was initially launched, we can plug in t = 0 into the equation h(t) = -4.9t² + 232t + 185.

h(0) = -4.9(0)² + 232(0) + 185

     = 0 + 0 + 185

     = 185

Therefore, the rocket was initially launched at a height of 185 meters.

To find the height of the rocket after 9 seconds, we can plug in t = 9 into the equation h(t) = -4.9t² + 232t + 185.

h(9) = -4.9(9)² + 232(9) + 185

     = -4.9(81) + 2088 + 185

     = -396.9 + 2088 + 185

     = 1876.1

Therefore, the rocket is at a height of 1876.1 meters after 9 seconds.

To find the velocity of the rocket after 9 seconds, we can take the derivative of the height function h(t) with respect to time (t) and evaluate it at t = 9.

The velocity function v(t) is the derivative of h(t) with respect to t:

v(t) = dh/dt = d/dt(-4.9t² + 232t + 185)

       = -9.8t + 232

v(9) = -9.8(9) + 232

       = -88.2 + 232

       = 143.8

Therefore, the velocity of the rocket after 9 seconds is 143.8 m/s.

To find the acceleration of the rocket after 9 seconds, we can take the derivative of the velocity function v(t) with respect to time (t) and evaluate it at t = 9.

The acceleration function a(t) is the derivative of v(t) with respect to t:

a(t) = dv/dt = d/dt(-9.8t + 232)

       = -9.8

a(9) = -9.8

Therefore, the acceleration of the rocket after 9 seconds is -9.8 m/s².

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