Question 1136201
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? 
When splash down occurs, h(t) = 0, therefore
-4.9t^2 + 169t + 234 = 0
find t using the quadratic formula; a=-4.9, b=169, c=234
 {{{t = (-169 +- sqrt(169^2-4*-4.9*234 ))/(2*-4.9) }}}
do the math, I a positive solution of
t = 35.8229
The rocket splashes down after 35.8229 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 max occurs on the axis of symmetry, use x=-b/(2a) to find it
t = {{{(-169)/(2*-4.9)}}}
t = 17.2449 sec
replace t with 17.2449 in the equation
h(t) = -4.9(17.2449^2) + 169(17.2449) + 234
do the math, I got
1691.194
The rocket peaks at 1691.194 meters above sea-level. 
graphically
{{{ graph( 300, 200, -10, 50, -1000, 2000, -4.9x^2+169x+234, 1691) }}}