Question

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)=−4.9t^2+169t+234.
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
The rocket splashes down after
seconds.[Round your answer to 4 decimal places)
How high above sea-level does the rocket get at its peak? [Round your answer to 3 decimal places]
The rocket peaks at
meters above sea-level.

Answer 1

Explanation:

To find the time at which the rocket will splash down, we need to set h(t) to 0 and solve for t:

h(t) = -4.9t^2 + 169t + 234 = 0

Using the quadratic formula, we get:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

Where a = -4.9, b = 169, and c = 234.

Plugging in the values, we get:

t = (-169 ± sqrt(169^2 - 4(-4.9)(234))) / 2(-4.9)

t = (-169 ± sqrt(28561.6)) / (-9.8)

We only want the positive value of t, as the rocket is launched at t = 0 seconds, so:

t = (-169 + sqrt(28561.6)) / (-9.8) = 17.2832 seconds (rounded to 4 decimal places)

Therefore, the rocket will splash down after 17.2832 seconds.

To find the height of the rocket at its peak, we need to find the vertex of the parabolic function h(t), which occurs at the value of t given by:

t = -b / 2a = -169 / 2(-4.9) = 17.3469 seconds (rounded to 4 decimal places)

At the time t = 17.3469 seconds, the height of the rocket is given by:

h(17.3469) = -4.9(17.3469)^2 + 169(17.3469) + 234 = 1764.203 meters (rounded to 3 decimal places)

Therefore, the rocket peaks at a height of 1764.203 meters above sea-level.