Question

NASA launches a rocket at t = 0 seconds. Its height, in meters above sea level, as a function of time is

Answer 1

Answer:

Step-by-step explanation:

y = ax² + bx + c

D = b² - 4ac

$x_{12}$ = ( - b ± √D ) ÷ 2a

$y_{max}$ = - $\frac{b}{2a}$

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

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( $-\frac{139}{2(-4.9)}$ )² + 139( $-\frac{139}{2(-4.9)}$ ) + 185 ≈ 1170.7653 ≈ 1171 meters

Answer 2

Answer:

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