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) = – 4.9t2 + 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.

Answer 1

Explanation:

y = ax² + bx + c

D = b² - 4ac

`x_(12)` = ( - b ± √D ) ÷ 2a

`y_(max)` = -
`(b)/(2a)`

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

h(t) = - 4.9t² + 139t + 185

- 4.9t² + 139t + 185 = 0

D = 139² + 4(- 4.9)(185) = 22947

`t_(12)` = ( - 139 ± √22947) ÷ 2(- 4.9)

`t_(1)` ≈ 29.641 ≈ 30 seconds

`t_(2)` < 0

`h_(max)` = - 4.9(
`-(139)/(2(-4.9))` )² + 139(
`-(139)/(2(-4.9))` ) + 185 ≈ 1170.7653 ≈ 1171 meters

Answer 2

the splash down occurs when h(t) = 0

0 = -4,9t2 + 139t + 185

using the quadratic formula,

t = -1.27 or 29.64 seconds

since time can't be negative

t = 29.64 secs

at the peak, h(t) is maximum

applying calculus,

[d{h(t)}}/dt = -9.8t + 139

at h(t) maximum, [d{h(t)}}/dt is 0

0 = -9.8t + 139

9.8t = 139

t = 14.18 secs

substituting,

h(t) = – 4.9t2 + 139t + 185

h(t) = -4.9(14.18^2) + 139(14.18) + 185

= 1170.77 m