Answer:
[tex] \displaystyle 1)48.2 \: \: \text{sec}[/tex]
[tex] \rm \displaystyle 2)3021.6 \: m[/tex]
Step-by-step explanation:
so when flash down occurs the rocket will be in the ground in other words the elevation(height) from ground level will be 0 therefore,
to figure out the time of flash down we can set h(t) to 0 by doing so we obtain:
[tex] \displaystyle - 4.9 {t}^{2} + 229t + 346 = 0[/tex]
to solve the equation can consider the quadratic formula given by
[tex] \displaystyle x = \frac{ - b \pm \sqrt{ {b}^{2} - 4 ac} }{2a} [/tex]
so let our a,b and c be -4.9,229 and 346 Thus substitute:
[tex] \rm\displaystyle t = \frac{ - (229) \pm \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)} [/tex]
remove parentheses:
[tex] \rm\displaystyle t = \frac{ - 229 \pm \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)} [/tex]
simplify square:
[tex] \rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 52441- 4( - 4.9)(346)} }{2.( - 4.9)} [/tex]
simplify multiplication:
[tex] \rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 52441- 6781.6} }{ - 9.8} [/tex]
simplify Substraction:
[tex] \rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 45659.4} }{ - 9.8} [/tex]
by simplifying we acquire:
[tex] \displaystyle t = 48.2 \: \: \: \text{and} \quad - 1.5[/tex]
since time can't be negative
[tex] \displaystyle t = 48.2 [/tex]
hence,
at 48.2 seconds splashdown occurs
to figure out the maximum height we have to figure out the maximum Time first in that case the following formula can be considered
[tex] \displaystyle x _{ \text{max}} = \frac{ - b}{2a} [/tex]
let a and b be -4.9 and 229 respectively thus substitute:
[tex] \displaystyle t _{ \text{max}} = \frac{ - 229}{2( - 4.9)} [/tex]
simplify which yields:
[tex] \displaystyle t _{ \text{max}} = 23.4[/tex]
now plug in the maximum t to the function:
[tex] \rm \displaystyle h(23.4)- 4.9 {(23.4)}^{2} + 229(23.4)+ 346 [/tex]
simplify:
[tex] \rm \displaystyle h(23.4) = 3021.6[/tex]
hence,
about 3021.6 meters high above sea-level the rocket gets at its peak?
